RuukkilabraUpper secondary physics
FY6 · at the end of the module

Apply independently

In these tasks nobody tells you which formula to use. Decide yourself which phenomena and models belong to the situation, solve the task on paper and justify your solution, as in the matriculation examination.

The four tasks bring the module together: the measurement of a battery, an electric circuit, an electric field and a capacitor. In each one you decide which models suit the situation, and you justify your choice.

Solve the task yourself first, on paper or on a computer. You may use a calculator, a data booklet and graphing software. Write out your reasoning, formulas, substitutions and answers with units.

Flashlight battery 15 points

Made-up example data · The measurement results have been made up for the task, but they are of the same order of magnitude as in a measurement of a new flat flashlight battery.

A student investigates a new flat flashlight battery marked $4.5\,\text{V}$ on its side. She connects an adjustable resistor and an ammeter to the battery, and a voltmeter directly across the terminals of the battery (figure). She adjusts the current with the resistor and reads the corresponding terminal voltage $U$. The meters show the current to the nearest $0.01\,\text{A}$ and the voltage to the nearest $0.01\,\text{V}$. The student measured the current in increasing order, and the circuit was connected for about a minute for each reading. The battery warmed up slightly during the measurements, and the voltage seemed to fall slowly even when the current was not changed. After the series the student measured the voltage once more at a current of $0.10\,\text{A}$: the reading was $4.45\,\text{V}$, that is $0.03\,\text{V}$ lower than at the beginning of the series.

Measurement results (current $I$, terminal voltage $U$):

  • $I = 0.10\,\text{A}$, $U = 4.48\,\text{V}$
  • $I = 0.20\,\text{A}$, $U = 4.37\,\text{V}$
  • $I = 0.30\,\text{A}$, $U = 4.22\,\text{V}$
  • $I = 0.40\,\text{A}$, $U = 4.13\,\text{V}$
  • $I = 0.50\,\text{A}$, $U = 3.99\,\text{V}$
  • $I = 0.60\,\text{A}$, $U = 3.89\,\text{V}$
$+$
$-$
$\text{V}$
$U$
$\text{A}$
$R$
$I$
  1. a) The student wants to find out what kind of voltage source the battery is. Model the battery and draw a graph from the data so that, according to the model, a straight line fits the points. 4 p
  2. b) Determine from the graph the source voltage $E$ and the internal resistance $R_\text{int}$ of the battery. 4 p
  3. c) An $8.0\,\Omega$ resistor is connected to the battery. How much power is transferred to the resistor, and what fraction of the whole electric power of the battery is transferred as heat in the internal resistance of the battery? 3 p
  4. d) Evaluate how the discharging and warming of the battery affect the result of part b. Also estimate how much uncertainty the scatter of the points leaves in the value of $R_\text{int}$ (the points deviate from the fitted line by at most $0.02\,\text{V}$), and suggest an improvement to the measurement. 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

a graph so that, according to the model of the battery, a straight line fits the points

Graph

The battery is modelled as an ideal voltage source with source voltage $E$ and an internal resistance $R_\text{int}$ in series with it. When a current $I$ flows through the battery, the voltage across the internal resistance is $R_\text{int} I$ and the terminal voltage is $U = E - R_\text{int} I$. This is the equation of a straight line: when the current $I$ is on the horizontal axis and the terminal voltage $U$ on the vertical axis, the line meets the vertical axis at $E$ and its slope is $-R_\text{int}$. So the measurements are plotted as they are.

$I\ (\text{A})$ $U\ (\text{V})$
$0.10$ $4.48$
$0.20$ $4.37$
$0.30$ $4.22$
$0.40$ $4.13$
$0.50$ $3.99$
$0.60$ $3.89$

The scale of the vertical axis starts at $3.6\,\text{V}$ so that the small decrease of the points can be seen, and the start is marked in the figure.

$0.1$
$0.2$
$0.3$
$0.4$
$0.5$
$0.6$
$0.7$
$0.8$
$3.8$
$4.0$
$4.2$
$4.4$
$4.6$
$4.8$
$I\ (\text{A})$
$U\ (\text{V})$
$3.6$
$0$
$\Delta I = 0.50\,\text{A}$
$\Delta U = -0.60\,\text{V}$
$E = 4.60\,\text{V}$

Model answer

The six points lie almost on the same straight line, so the model $U = E - R_\text{int} I$ describes the behaviour of the battery well in the measured range: the terminal voltage decreases steadily as the current increases. The line has been fitted between the points, and the points have not been joined into a broken line. The line has been extended as a dashed line to the vertical axis so that the point $I = 0$, the source voltage, can be seen.

b)

Given

the line of part a, from which the points $(0.10\,\text{A}$; $4.48\,\text{V})$ and $(0.60\,\text{A}$; $3.88\,\text{V})$ are read

Find

the source voltage $E$ and the internal resistance $R_\text{int}$

Principle

According to the model the terminal voltage is $U = E - R_\text{int} I$. The point where the line meets the vertical axis is at current zero, when there is no voltage drop inside the battery, so the intercept is $E$. The slope tells how much the terminal voltage falls as the current increases, so it is $-R_\text{int}$.

Reading the graph

The slope is read from two points of the drawn line that are far apart, not from adjacent measurement points: $(0.10\,\text{A}$; $4.48\,\text{V})$ and $(0.60\,\text{A}$; $3.88\,\text{V})$. Between them $\Delta I = 0.50\,\text{A}$ and $\Delta U = -0.60\,\text{V}$ (the slope triangle of the graph). The line meets the vertical axis at $4.60\,\text{V}$.

Formula

$R_\text{int}$$= -\dfrac{\Delta U}{\Delta I}$

$E = U + R_\text{int} I$ from any point of the line.

Conversions

Not needed: the voltage is in volts and the current in amperes, and $1\,\text{V/A} = 1\,\Omega$.

Substitution

$R_\text{int}$$= -\dfrac{-0.60\,\text{V}}{0.50\,\text{A}}$
$= 1.2\,\Omega$
$E$$= 4.48\,\text{V} + 1.2\,\Omega \cdot 0.10\,\text{A}$
$= 4.48\,\text{V} + 0.12\,\text{V}$
$= 4.60\,\text{V}$

Answer

$E \approx 4.6\,\text{V}$

And $R_\text{int} \approx 1.2\,\Omega$, two significant figures, because the slope is read from a graph and cannot be read more precisely than to two figures.

Reasonableness

The source voltage is slightly larger than the rated voltage of the battery, $4.5\,\text{V}$, which suits a new battery. The internal resistance is of the order of an ohm, as in flat flashlight batteries. The short-circuit current would be

$\dfrac{E}{R_\text{int}}$$\approx 3.8\,\text{A}$

c)

Given

$E = 4.60\,\text{V}$, $R_\text{int} = 1.2\,\Omega$ and the external resistor $R = 8.0\,\Omega$

Find

the power $P$ transferred to the resistor and the fraction of the whole electric power of the battery that is transferred as heat in the internal resistance of the battery

Principle

In the circuit the battery, the internal resistance and the resistor are in series, so the same current flows through all of them. According to the loop rule the source voltage is divided between the internal resistance and the resistor: $E - R_\text{int} I - R I = 0$. Of the electric power, heat is produced both in the resistor and inside the battery, and the whole power of the battery is the sum of these, $E I$.

Formula

$I = \dfrac{E}{R + R_\text{int}}$, $P = R I^2$, $P_\text{int} = R_\text{int} I^2$ and the fraction

$\dfrac{P_\text{int}}{P + P_\text{int}}$$= \dfrac{R_\text{int}}{R + R_\text{int}}$

Conversions

Not needed, all quantities are in base units.

Substitution

$I$$= \dfrac{4.60\,\text{V}}{8.0\,\Omega + 1.2\,\Omega}$
$= \dfrac{4.60\,\text{V}}{9.2\,\Omega}$
$= 0.50\,\text{A}$
$P$$= 8.0\,\Omega \cdot (0.50\,\text{A})^2$
$= 2.0\,\text{W}$
$P_\text{int}$$= 1.2\,\Omega \cdot (0.50\,\text{A})^2$
$= 0.30\,\text{W}$

The fraction is

$\dfrac{0.30\,\text{W}}{2.0\,\text{W} + 0.30\,\text{W}}$$= 0.13$

Answer

The power transferred to the resistor is $P \approx 2.0\,\text{W}$, and about $13\,\%$ of the electric power of the battery is transferred as heat inside the battery. The result is given to two significant figures like the initial data.

Reasonableness

The terminal voltage is $R I = 4.0\,\text{V}$, which is the same as $E - R_\text{int} I = 4.60\,\text{V} - 0.60\,\text{V}$, and on the graph the voltage at a current of $0.50\,\text{A}$ is about $4.0\,\text{V}$. Since the external resistance is more than seven times the internal resistance, only a small part of the power stays inside the battery.

d)

Find

the effect of discharging and warming on the result for the source voltage and the internal resistance, the uncertainty of $R_\text{int}$ caused by the scatter, and an improvement of the measurement

Principle

The model $U = E - R_\text{int} I$ assumes that $E$ and $R_\text{int}$ stay constant throughout the measurement. If the battery discharges or warms up, they change during the measurement, and the line of the graph describes a kind of average. The scatter tells about random errors, that is, how much the fitted line could turn.

Model answer

Systematic error. The current was increased all the time in the same direction, and for each reading the circuit was connected for about a minute. The discharging of the battery lowers the voltage over time, so the voltages measured later, that is, those corresponding to the largest currents, are too small. Since time and current increase together, the effect of time is mixed with the slope: the line falls too steeply and $R_\text{int}$ comes out too large. Warming also changes $R_\text{int}$ during the measurement (usually it makes it smaller), so the direction of its effect is uncertain. The effect on the source voltage is small, because the smallest current was measured first with a fresh battery.

The size of the drift. The new measurement at a current of $0.10\,\text{A}$ gave $4.45\,\text{V}$, that is $0.03\,\text{V}$ less than at the beginning of the series. At worst the last point is therefore about $0.03\,\text{V}$ too low, and the slope is too large by at most $\dfrac{0.03\,\text{V}}{0.50\,\text{A}} = 0.06\,\Omega$.

Uncertainty. The points deviate from the fitted line by at most $0.02\,\text{V}$. If one end of the line rises and the other falls by this much, the slope changes by $\dfrac{2 \cdot 0.02\,\text{V}}{0.50\,\text{A}} = 0.08\,\Omega$, that is, $R_\text{int} = (1.20 \pm 0.08)\,\Omega$. The uncertainty has been rounded to one significant figure.

Improvement. The currents are measured in random order, or the series is repeated in reverse order, and the circuit is connected only for the time of the reading. If the results differ from each other, the difference tells the size of the effect of discharging.

Conclusion. The effect of the drift ($0.06\,\Omega$) is of the same order of magnitude as that of the scatter ($0.08\,\Omega$), so $R_\text{int}$ is about $1.1\ldots1.2\,\Omega$. The value $1.2\,\Omega$ is rather an upper limit, because the systematic error increases it.

Also accepted

  • In part a the vertical axis of the graph may start from zero as long as the points and $E$ can be distinguished; then the line may also be extended to the horizontal axis, where $I_\text{sc} = \dfrac{E}{R_\text{int}} \approx 3.8\,\text{A}$.
  • In part b a slope reading of $-1.1\ldots-1.3\,\text{V/A}$ (answer $R_\text{int}$ about $1.1\ldots1.3\,\Omega$) and a source voltage of $4.5\ldots4.7\,\text{V}$ are accepted. A fit made with a calculator or a spreadsheet is also accepted.
  • In part d the uncertainty may also be estimated by fitting the extreme lines and reading their slopes.

Common mistakes

  • The points are joined into a broken line, or the line is forced through the origin even though $U$ is not zero when the current is zero.
  • The slope is read from two consecutive measurement points instead of points of the line that are far apart.
  • The sign of the slope is ignored, or $R_\text{int}$ is calculated as $U : I$ from a single measurement, which gives the value of the external resistance, not the internal resistance.
  • In part c the whole source voltage $E$ is taken as the voltage of the resistor and the voltage drop in the internal resistance is forgotten.
  • The uncertainty is written as the reading precision of the meter, $0.01\,\text{V}$, as such, without taking into account the scatter of the points or the systematic error.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)4 p

Models the battery as an ideal voltage source with an internal resistance in series with it, so that the terminal voltage is $U = E - R_\text{int} I$, and chooses $I$ and $U$ as the quantities of the graph. 1 p
The axes show the correct quantities and units, and the scales are evenly spaced. $I\ (\text{A})$ on the horizontal axis and $U\ (\text{V})$ on the vertical axis; the vertical axis does not have to start from zero if the scale is marked. 1 p
Each measurement is marked as a separate point in the right place. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. The line passes between the points and has been extended to the vertical axis. 1 p

b)4 p

The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The slope $\dfrac{\Delta U}{\Delta I} \approx -1.2\,\text{V/A}$ has been read from two points of the drawn line that are far apart, for example $(0.10\,\text{A}$; $4.48\,\text{V})$ and $(0.60\,\text{A}$; $3.88\,\text{V})$, not from adjacent measurement points. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $U = E - R_\text{int} I$: $E$ is the point where the line meets the vertical axis ($I = 0$) and $R_\text{int}$ is the negative of the slope. 1 p
The numerical values are substituted with their units, and the calculation is correct. $R_\text{int} = \dfrac{0.60\,\text{V}}{0.50\,\text{A}}$ and $E = 4.48\,\text{V} + R_\text{int} \cdot 0.10\,\text{A}$, or the intercept read from the line. 1 p
The answer has the correct unit and a precision that matches the given data. $E \approx 4.6\,\text{V}$ and $R_\text{int} \approx 1.2\,\Omega$ 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $I = \dfrac{E}{R + R_\text{int}}$, the power of the resistor $P = R I^2$ and inside the battery $P_\text{int} = R_\text{int} I^2$; the fraction is $\dfrac{P_\text{int}}{P + P_\text{int}}$. 1 p
The numerical values are substituted with their units, and the calculation is correct. $I = 0.50\,\text{A}$, $P = 2.0\,\text{W}$ and $P_\text{int} = 0.30\,\text{W}$. 1 p
The answer has the correct unit and a precision that matches the given data. $P \approx 2.0\,\text{W}$ and the fraction about $13\,\%$ 1 p

d)4 p

The main sources of error are named, and it is explained how they affect the result. The discharging of the battery lowers the voltage during the series, so the voltages measured later (at larger currents) are too small: the line is too steep and $R_\text{int}$ comes out too large (a systematic error). Because the current was only increased in one direction, the effect of time is mixed with the slope. Warming also changes $R_\text{int}$ during the measurement (usually it makes it smaller), so its direction is uncertain. 1 p
The uncertainty of the measurement has been estimated. The slopes of the extreme lines differ by $\dfrac{2 \cdot 0.02\,\text{V}}{0.50\,\text{A}} = 0.08\,\Omega$, so on the basis of the scatter alone $R_\text{int} = (1.20 \pm 0.08)\,\Omega$. 1 p
Suggests an improvement that reduces the effect of discharging: the currents in random order, the circuit connected only for the time of the reading, or the series repeated in reverse order and the results compared. 1 p
The conclusion answers the question and is based on the data. Estimates the effect of the drift: the voltage fell by about $0.03\,\text{V}$ during the series, so the slope is too large by at most about $\dfrac{0.03\,\text{V}}{0.50\,\text{A}} = 0.06\,\Omega$. This is of the same order of magnitude as the scatter $0.08\,\Omega$, so $R_\text{int}$ is about $1.1\ldots1.2\,\Omega$ and the value $1.2\,\Omega$ is rather an upper limit. 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Cabin yard lights 15 points

Made-up example data · The resistance of the cable and the data of the lamps are made up, but of the same order of magnitude as in a 12 volt cabin electrical system.

The yard lights of a cabin get their electricity from a $12.0\,\text{V}$ battery bank. A long cable leads from the battery bank to the lamps, and the combined resistance of both of its conductors is $0.80\,\Omega$. After the battery bank there is a fuse in the circuit. At the end of the cable there are two lamps connected in parallel (figure). Lamp A is marked $12\,\text{V},\ 24\,\text{W}$ and lamp B $12\,\text{V},\ 12\,\text{W}$: the marking gives the power when the rated voltage is connected to the lamp. Assume that the resistances of the lamps are constant. The internal resistance of the battery bank is so small that it can be ignored.

Other data:

  • the rated current of the fuse is $3.15\,\text{A}$
  • the cable can withstand a current of at most $4.0\,\text{A}$ for a long time
$+$
$-$
$E$
$\text{fuse}$
$R_\text{k}$
$\text{A}$
$12\,\text{V}$
$24\,\text{W}$
$\text{B}$
$12\,\text{V}$
$12\,\text{W}$
$I$
  1. a) How large is the current from the battery bank when both lamps are lit? 4 p
  2. b) How much power is actually transferred to lamp A? Compare the result with the rated power of the lamp. 3 p
  3. c) Lamp B burns out and no longer conducts current. How do the brightness of lamp A, the current from the battery bank and the warming of the cable change? Justify your answer, and check your reasoning by calculating the power of lamp A. 4 p
  4. d) The owner adds a third lamp C in parallel to the circuit, marked $12\,\text{V},\ 36\,\text{W}$. How large a current will now flow from the battery bank? What happens to the fuse, and why is the rated current of the fuse chosen according to the cable? Also calculate the voltage of the lamps. 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

source voltage $E = 12.0\,\text{V}$, resistance of the cable $R_\text{c} = 0.80\,\Omega$, lamp A $12\,\text{V},\ 24\,\text{W}$ and lamp B $12\,\text{V},\ 12\,\text{W}$

Find

the current $I$ from the battery bank when both lamps are lit

Principle

A lamp is treated as a constant resistance whose value is obtained from the rated values: at the rated voltage the power is $P = \dfrac{U^2}{R}$. Lamps A and B are in parallel, so they have the same voltage, and the cable is in series with them, so the whole current flows through the cable. Then the total resistance is the sum of the resistance of the cable and the equivalent resistance of the parallel connection.

Formula

$R_\text{A}$$= \dfrac{U_\text{n}^2}{P_\text{A}}$
$R_\text{B}$$= \dfrac{U_\text{n}^2}{P_\text{B}}$
$\dfrac{1}{R_\text{par}}$$= \dfrac{1}{R_\text{A}} + \dfrac{1}{R_\text{B}}$

$R = R_\text{c} + R_\text{par}$ and $I = \dfrac{E}{R}$

Conversions

Not needed: $1\,\text{V}^2/\text{W} = 1\,\Omega$.

Substitution

$R_\text{A}$$= \dfrac{(12\,\text{V})^2}{24\,\text{W}}$
$= 6.0\,\Omega$
$R_\text{B}$$= \dfrac{(12\,\text{V})^2}{12\,\text{W}}$
$= 12\,\Omega$

Parallel connection:

$\dfrac{1}{R_\text{par}}$$= \dfrac{1}{6.0\,\Omega} + \dfrac{1}{12\,\Omega}$
$= \dfrac{3}{12\,\Omega}$

so $R_\text{par} = 4.0\,\Omega$. The total resistance is

$R$$= 0.80\,\Omega + 4.0\,\Omega$
$= 4.8\,\Omega$

and the current

$I$$= \dfrac{12.0\,\text{V}}{4.8\,\Omega}$
$= 2.5\,\text{A}$

Answer

$I = 2.5\,\text{A}$

Two significant figures as in the initial data.

Reasonableness

If the lamps got the rated voltage, their currents would be $2.0\,\text{A}$ and $1.0\,\text{A}$, a total of $3.0\,\text{A}$. The cable takes part of the voltage, so the current is smaller than this. The current is also below the limits of the fuse, $3.15\,\text{A}$, and of the cable, $4.0\,\text{A}$.

b)

Given

$E = 12.0\,\text{V}$, $R_\text{c} = 0.80\,\Omega$, $R_\text{A} = 6.0\,\Omega$ and the current of part a, $I = 2.5\,\text{A}$

Find

the actual power $P_\text{A}$ of lamp A and its comparison with the rated power $24\,\text{W}$

Principle

According to the loop rule the voltage of the battery bank is divided between the voltage drop in the cable and the voltage of the lamps: $E - R_\text{c} I - U_\text{L} = 0$. The voltage of the lamps is therefore lower than the rated voltage, and the power is calculated from this actual voltage.

Formula

$U_\text{L}$$= E - R_\text{c} I$
$P_\text{A}$$= \dfrac{U_\text{L}^2}{R_\text{A}}$

Conversions

Not needed.

Substitution

$U_\text{L}$$= 12.0\,\text{V} - 0.80\,\Omega \cdot 2.5\,\text{A}$
$= 12.0\,\text{V} - 2.0\,\text{V}$
$= 10.0\,\text{V}$
$P_\text{A}$$= \dfrac{(10.0\,\text{V})^2}{6.0\,\Omega}$
$= 16.7\,\text{W}$

Answer

$P_\text{A} \approx 17\,\text{W}$

About $70\,\%$ of the rated $24\,\text{W}$, two significant figures as in the initial data ($0.80\,\Omega$, $24\,\text{W}$, $12\,\text{V}$).

Reasonableness

The lamp shines dimmer than at the rated voltage, because its voltage is only $10.0\,\text{V}$. Check: the power of lamp B is

$\dfrac{(10.0\,\text{V})^2}{12\,\Omega}$$= 8.3\,\text{W}$

and the heat transferred to the cable is $0.80\,\Omega \cdot (2.5\,\text{A})^2 = 5.0\,\text{W}$. The sum $16.7\,\text{W} + 8.3\,\text{W} + 5.0\,\text{W} = 30\,\text{W}$ is equal to the power of the battery bank, $12.0\,\text{V} \cdot 2.5\,\text{A} = 30\,\text{W}$.

c)

Find

how the brightness of lamp A, the current from the battery bank and the warming of the cable change when lamp B is removed from the circuit, and a check of the power of lamp A by calculation

Principle

The battery bank gives the circuit a constant voltage of $12.0\,\text{V}$. When one parallel branch is removed, the equivalent resistance of the parallel connection increases and the total resistance increases, so the current decreases. The cable is in series with lamp A, so its voltage drop depends on the whole current.

Model answer

Current. The total resistance is now $R = 0.80\,\Omega + 6.0\,\Omega = 6.8\,\Omega$ and the current $I = \dfrac{12.0\,\text{V}}{6.8\,\Omega} \approx 1.8\,\text{A}$, previously $2.5\,\text{A}$.

Brightness. The smaller current reduces the voltage drop in the cable: $0.80\,\Omega \cdot 1.76\,\text{A} \approx 1.4\,\text{V}$, previously $2.0\,\text{V}$. A larger voltage $U_\text{A} \approx 10.6\,\text{V}$ is therefore left for lamp A, and its power is $\dfrac{(10.6\,\text{V})^2}{6.0\,\Omega} \approx 18.7\,\text{W}$, previously $16.7\,\text{W}$. So lamp A shines brighter, although nothing changed in its own branch.

Cable. The power transferred to the cable as heat is $R_\text{c} I^2 = 0.80\,\Omega \cdot (1.76\,\text{A})^2 \approx 2.5\,\text{W}$, previously $5.0\,\text{W}$. So the cable warms up less, because the power depends on the square of the current.

Conclusion. Lamps connected in parallel are independent of each other only if the voltage across them stays constant. The resistance of the cable makes the lamps slightly dependent on each other: if the resistance of the cable were zero, the brightness of lamp A would not change at all.

d)

Given

$E = 12.0\,\text{V}$, $R_\text{c} = 0.80\,\Omega$, $R_\text{A} = 6.0\,\Omega$, $R_\text{B} = 12\,\Omega$, new lamp C $12\,\text{V},\ 36\,\text{W}$, fuse $3.15\,\text{A}$ and cable limit $4.0\,\text{A}$

Find

the current $I$ from the battery bank with three lamps and the operation of the fuse

Principle

The lamps are still in parallel, so they have a common voltage and the resistances are added as reciprocals. The cable is in series with all of them. The fuse is a protective device of the circuit: it blows and breaks the circuit if the current exceeds its rated current for too long.

Formula

$R_\text{C}$$= \dfrac{U_\text{n}^2}{P_\text{C}}$
$\dfrac{1}{R_\text{par}}$$= \dfrac{1}{R_\text{A}} + \dfrac{1}{R_\text{B}} + \dfrac{1}{R_\text{C}}$
$I$$= \dfrac{E}{R_\text{c} + R_\text{par}}$

Conversions

Not needed.

Substitution

$R_\text{C}$$= \dfrac{(12\,\text{V})^2}{36\,\text{W}}$
$= 4.0\,\Omega$
$\dfrac{1}{R_\text{par}}$$= \dfrac{1}{6.0\,\Omega} + \dfrac{1}{12\,\Omega} + \dfrac{1}{4.0\,\Omega}$
$= \dfrac{2 + 1 + 3}{12\,\Omega}$
$= \dfrac{1}{2.0\,\Omega}$

so $R_\text{par} = 2.0\,\Omega$. The current is

$I$$= \dfrac{12.0\,\text{V}}{0.80\,\Omega + 2.0\,\Omega}$
$= \dfrac{12.0\,\text{V}}{2.8\,\Omega}$
$= 4.3\,\text{A}$

Answer

$I \approx 4.3\,\text{A}$

The current exceeds both the rated current of the fuse, $3.15\,\text{A}$, and the permitted current of the cable, $4.0\,\text{A}$, so the fuse will probably blow if the overcurrent continues and break the circuit (the actual operating time depends on the type of the fuse, so this is a simplification). The voltage of the lamps is $R_\text{par} I = 2.0\,\Omega \cdot 4.3\,\text{A} \approx 8.6\,\text{V}$, so they shine dimmer than before. The fuse is chosen to protect the weakest part of the circuit, the cable: its rated current is lower than the permitted current of the cable, so that the cable does not warm up dangerously before the circuit is broken.

Reasonableness

The third lamp reduces the resistance of the parallel connection from $4.0\,\Omega$ to $2.0\,\Omega$, so the current must increase from $2.5\,\text{A}$. Without a fuse, $0.80\,\Omega \cdot (4.3\,\text{A})^2 \approx 15\,\text{W}$ would be transferred to the cable as heat, almost three times the earlier $5.0\,\text{W}$. The resistances of the lamps were calculated from the rated values, so the result is an estimate.

Also accepted

  • Part a may be calculated from the rated currents of the lamps $I = \dfrac{P}{U}$ and the resistances $R = \dfrac{U}{I}$, as long as the results are the same: $2.0\,\text{A}$ and $1.0\,\text{A}$, $6.0\,\Omega$ and $12\,\Omega$.
  • In part b the voltage of the lamp may also be calculated directly as $U_\text{L} = R_\text{par} I = 4.0\,\Omega \cdot 2.5\,\text{A}$.
  • In part d the voltage of the lamps ($8.6\,\text{V}$) may also be calculated and it may be stated that the lamps shine dimmer than before.

Common mistakes

  • The resistances of the lamps are calculated with the formula $R = \dfrac{P}{U}$ or $R = \dfrac{U}{P}$.
  • The resistance of the parallel connection is calculated as the sum $6.0\,\Omega + 12\,\Omega$, or the reciprocal obtained is not converted back ($\tfrac{1}{4.0\,\Omega} = 0.25\,\text{S}$ is written as a resistance $0.25\,\Omega$).
  • The whole voltage of the battery bank, $12.0\,\text{V}$, is taken as the voltage of the lamps, and the voltage drop in the cable is forgotten.
  • In part c it is concluded that lamp A does not change at all because it is in parallel: it is forgotten that the cable is in series.
  • In part d it is noted that the fuse is exceeded, but the purpose of the fuse as a protection of the cable is not explained.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)4 p

Sees the structure of the circuit: the cable is in series with the lamps connected in parallel. Calculates the resistances of the lamps from the rated values, for example $R = \dfrac{U^2}{P}$: $R_\text{A} = 6.0\,\Omega$ and $R_\text{B} = 12\,\Omega$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\dfrac{1}{R_\text{par}} = \dfrac{1}{R_\text{A}} + \dfrac{1}{R_\text{B}}$, $R = R_\text{c} + R_\text{par}$ and $I = \dfrac{E}{R}$ solved before the numerical values. 1 p
The numerical values are substituted with their units, and the calculation is correct. $R_\text{par} = 4.0\,\Omega$ and $R = 4.8\,\Omega$. 1 p
The answer has the correct unit and a precision that matches the given data. $I = 2.5\,\text{A}$ 1 p

b)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. The loop rule $E - R_\text{c} I - U_\text{L} = 0$ gives the voltage of the lamps $U_\text{L} = E - R_\text{c} I$; the power of the lamp is $P = \dfrac{U_\text{L}^2}{R_\text{A}}$. 1 p
The numerical values are substituted with their units, and the calculation is correct. $U_\text{L} = 12.0\,\text{V} - 0.80\,\Omega \cdot 2.5\,\text{A} = 10.0\,\text{V}$ and $P_\text{A} = \dfrac{(10.0\,\text{V})^2}{6.0\,\Omega} = 16.7\,\text{W}$. 1 p
The reasonableness of the result has been assessed. $17\,\text{W}$ is less than the rated $24\,\text{W}$, because the voltage drop in the cable leaves less than $12\,\text{V}$ for the lamp. Check: the power of the battery bank, $30\,\text{W}$, is $16.7\,\text{W} + 8.3\,\text{W} + 5.0\,\text{W}$. 1 p

c)4 p

Concludes that the total resistance increases when one parallel branch is removed, so the current from the battery bank decreases: $I = \dfrac{12.0\,\text{V}}{6.8\,\Omega} \approx 1.8\,\text{A}$. 1 p
Justifies that a smaller current reduces the voltage drop in the cable ($R_\text{c} I$ decreases), so a larger voltage is left for lamp A and it shines brighter, and that the power transferred to the cable, $R_\text{c} I^2$, decreases from about $5.0\,\text{W}$ to $2.5\,\text{W}$, so the cable warms up less. 1 p
Calculates the power of lamp A: $U_\text{A} \approx 10.6\,\text{V}$ and $P_\text{A} \approx 18.7\,\text{W}$, previously $16.7\,\text{W}$. 1 p
The conclusion answers the question and is based on the data. States that lamps connected in parallel are not completely independent of each other when the cable has resistance: if the resistance of the cable were zero, the brightness of lamp A would not change. 1 p

d)4 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. Three parallel lamps: $\dfrac{1}{R_\text{par}} = \dfrac{1}{R_\text{A}} + \dfrac{1}{R_\text{B}} + \dfrac{1}{R_\text{C}}$ and $I = \dfrac{E}{R_\text{c} + R_\text{par}}$. 1 p
The numerical values are substituted with their units, and the calculation is correct. $R_\text{C} = \dfrac{(12\,\text{V})^2}{36\,\text{W}} = 4.0\,\Omega$, $R_\text{par} = 2.0\,\Omega$ and $I = \dfrac{12.0\,\text{V}}{2.8\,\Omega} = 4.3\,\text{A}$. 1 p
The conclusion answers the question and is based on the data. The current $4.3\,\text{A}$ exceeds the $3.15\,\text{A}$ of the fuse and the permitted $4.0\,\text{A}$ of the cable, so the fuse will probably blow if the overcurrent continues and break the circuit; the operating time of the fuse is simplified, since a small overcurrent can last for a moment. The fuse is chosen to protect the weakest part, the cable, so its rated current is lower than the permitted current of the cable. 1 p
The numerical values are substituted with their units, and the calculation is correct. The voltage of the lamps is $R_\text{par} I = 2.0\,\Omega \cdot 4.3\,\text{A} \approx 8.6\,\text{V}$, so the lamps shine dimmer than at a voltage of $10.0\,\text{V}$. 1 p

Self-assessed points: 0 / 15 (15 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Spray painting 16 points

Made-up example data · The orders of magnitude are of the same kind as in industrial electrostatic spray painting, but the values of the task have been rounded and simplified.

A car part is painted with an electrostatic spray gun. The potential of the nozzle plate is $+60\,\text{kV}$ and the grounded workpiece is $20\,\text{cm}$ away. The paint droplets leaving the nozzle get a positive charge, and they move towards the workpiece. Assume that the electric field between the nozzle plate and the workpiece is uniform (figure).

One paint droplet:

  • mass $4.0 \cdot 10^{-10}\,\text{kg}$
  • charge $q = +6.0 \cdot 10^{-13}\,\text{C}$
  • leaves the nozzle plate from rest

Other data: the dielectric strength of air is about $3\,\text{MV/m}$, the acceleration of free fall is $9.81\,\text{m/s}^2$ and the constant of Coulomb's law is $k = 8.988 \cdot 10^{9}\,\tfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$.

$+60\,\text{kV}$
$0\,\text{V}$
$\text{nozzle plate}$
$\text{workpiece}$
$\overline{E}$
$d$
$+$
$q$
  1. a) How large is the electric force acting on the droplet? Is it justified to ignore the weight of the droplet, and will sparking occur in the field? 4 p
  2. b) At what speed does the droplet hit the workpiece when air resistance is not taken into account? 4 p
  3. c) In a measurement the impact speed of the droplets is found to be about $5.0\,\text{m/s}$. Explain why it is much smaller than the result of part b, and conclude where the work done by the electric field goes. 4 p
  4. d) Two neighbouring droplets are $1.0\,\text{mm}$ apart, and the charge of each is $+6.0 \cdot 10^{-13}\,\text{C}$. How large is the force between the droplets? What does the result tell about the behaviour of the cloud of droplets? 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

potential difference $U = 60\,\text{kV} = 6.0 \cdot 10^{4}\,\text{V}$, distance $d = 20\,\text{cm}$, charge $q = 6.0 \cdot 10^{-13}\,\text{C}$ and mass $m = 4.0 \cdot 10^{-10}\,\text{kg}$

Find

the electric force $F$ acting on the droplet, the significance of the weight and the possibility of sparking

Principle

Between the plates there is a uniform electric field whose strength is the potential difference divided by the distance. A force in the direction of the field acts on the positively charged droplet, so the droplet moves towards the lower potential, that is, the grounded workpiece. The significance of the weight is assessed by comparing it with the electric force, and sparking by comparing the strength of the field with the dielectric strength of air.

Formula

$E = \dfrac{U}{d}$ and $F = qE$

Conversions

$20\,\text{cm}$$= 0.20\,\text{m}$
$60\,\text{kV}$$= 60 \cdot 1000\,\text{V}$
$= 6.0 \cdot 10^{4}\,\text{V}$

Substitution

$E$$= \dfrac{6.0 \cdot 10^{4}\,\text{V}}{0.20\,\text{m}}$
$= 3.0 \cdot 10^{5}\,\text{V/m}$
$F$$= 6.0 \cdot 10^{-13}\,\text{C} \cdot 3.0 \cdot 10^{5}\,\text{V/m}$
$= 1.8 \cdot 10^{-7}\,\text{N}$

Answer

$F \approx 1.8 \cdot 10^{-7}\,\text{N}$

In the direction of the field towards the workpiece, two significant figures as in the initial data.

Reasonableness

The weight of the droplet is

$mg$$= 4.0 \cdot 10^{-10}\,\text{kg} \cdot 9.81\,\text{m/s}^2$
$= 3.9 \cdot 10^{-9}\,\text{N}$

only about $2\,\%$ of the electric force, so the weight can be ignored. The strength of the field, $3.0 \cdot 10^{5}\,\text{V/m} = 0.30\,\text{MV/m}$, is about a tenth of the dielectric strength of air, $3\,\text{MV/m}$, so no sparking occurs.

b)

Given

$q = 6.0 \cdot 10^{-13}\,\text{C}$, $U = 6.0 \cdot 10^{4}\,\text{V}$, $m = 4.0 \cdot 10^{-10}\,\text{kg}$ and the droplet leaves from rest

Find

the speed $v$ of the droplet when it hits the workpiece, when air resistance is not taken into account

Principle

The work done by the electric field on the droplet is the charge times the potential difference, $W = qU$, and it increases the kinetic energy of the droplet. Weight and air resistance are left out, so there are no other forces. In this way it is enough to know the potential difference, and the strength of the field or the acceleration is not needed.

Formula

$q U = \dfrac{1}{2} m v^2$, so $v = \sqrt{\dfrac{2 q U}{m}}$

Conversions

Not needed, all quantities are in base units; $1\,\text{C} \cdot 1\,\text{V} = 1\,\text{J}$.

Substitution

$qU$$= 6.0 \cdot 10^{-13}\,\text{C} \cdot 6.0 \cdot 10^{4}\,\text{V}$
$= 3.6 \cdot 10^{-8}\,\text{J}$
$v$$= \sqrt{\dfrac{2 \cdot 3.6 \cdot 10^{-8}\,\text{J}}{4.0 \cdot 10^{-10}\,\text{kg}}}$
$= \sqrt{180\,\text{m}^2/\text{s}^2}$
$= 13.4\,\text{m/s}$

Answer

$v \approx 13\,\text{m/s}$

Two significant figures as in the initial data.

Reasonableness

The acceleration is

$\dfrac{F}{m}$$= \dfrac{1.8 \cdot 10^{-7}\,\text{N}}{4.0 \cdot 10^{-10}\,\text{kg}}$
$= 450\,\text{m/s}^2$

so over a distance of $0.20\,\text{m}$ the speed would be $\sqrt{2 \cdot 450\,\text{m/s}^2 \cdot 0.20\,\text{m}} = 13.4\,\text{m/s}$, which confirms the result. The work done by the weight over the whole distance would be only $7.8 \cdot 10^{-10}\,\text{J}$, about $2\,\%$ of the work done by the electric field.

c)

Find

why the measured impact speed $5.0\,\text{m/s}$ is much smaller than the result of part b and where the work done by the electric field goes

Principle

The model of part b left out air resistance. Air resistance acts against the direction of motion of the droplet, and the work it does is negative. When the speed of the droplet increases, air resistance increases, and the droplet moves at a speed at which air resistance and the electric force are approximately equal. According to the conservation of energy, the work done by the electric field is not lost but goes elsewhere.

Model answer

Explanation. The speed $13\,\text{m/s}$ of part b would be possible only in a vacuum. In air, air resistance limits the speed to a much smaller value, so the result of part b is the greatest possible speed of the droplet, an upper limit.

Calculation. The actual kinetic energy of the droplet is $\dfrac{1}{2} m v^2 = \dfrac{1}{2} \cdot 4.0 \cdot 10^{-10}\,\text{kg} \cdot (5.0\,\text{m/s})^2 = 5.0 \cdot 10^{-9}\,\text{J}$. The work done by the electric field is $3.6 \cdot 10^{-8}\,\text{J}$, so only $\dfrac{5.0 \cdot 10^{-9}\,\text{J}}{3.6 \cdot 10^{-8}\,\text{J}} \approx 14\,\%$ ends up as kinetic energy.

The fate of the energy. The rest, $3.6 \cdot 10^{-8}\,\text{J} - 5.0 \cdot 10^{-9}\,\text{J} = 3.1 \cdot 10^{-8}\,\text{J}$, is transferred to the air. The internal energy of the air increases, and the energy is transferred further as heat to the surroundings. So energy is not lost.

Conclusion. The assumption of part b (no air resistance) makes the speed too large, so the result is the greatest possible speed. Energy is not lost but is transferred to the air.

d)

Given

$q_1 = q_2 = 6.0 \cdot 10^{-13}\,\text{C}$, $r = 1.0\,\text{mm}$ and $k = 8.988 \cdot 10^{9}\,\tfrac{\text{N}\cdot\text{m}^2}{\text{C}^2}$

Find

the force $F$ between the droplets and its significance for the behaviour of the cloud of droplets

Principle

The droplets are much smaller than the distance between them, so they can be treated as point charges. The force between them obeys Coulomb's law: charges of the same sign repel each other.

Formula

$F = k\dfrac{q_1 q_2}{r^2}$

Conversions

$1.0\,\text{mm} = 1.0 \cdot 10^{-3}\,\text{m}$

Substitution

$F$$= 8.988 \cdot 10^{9}\,\tfrac{\text{N}\cdot\text{m}^2}{\text{C}^2} \cdot \dfrac{(6.0 \cdot 10^{-13}\,\text{C})^2}{(1.0 \cdot 10^{-3}\,\text{m})^2}$
$= 8.988 \cdot 10^{9} \cdot \dfrac{3.6 \cdot 10^{-25}}{1.0 \cdot 10^{-6}}\,\text{N}$
$= 3.2 \cdot 10^{-9}\,\text{N}$

Answer

$F \approx 3.2 \cdot 10^{-9}\,\text{N}$

Repulsive, two significant figures as in the initial data.

Reasonableness

The force is about $1.8\,\%$ of the force of the field, $1.8 \cdot 10^{-7}\,\text{N}$, but $0.8$ times the weight of the droplet, $3.9 \cdot 10^{-9}\,\text{N}$. The repulsive force is therefore small compared with the force of the electric field, so the droplets still move towards the workpiece, but the repulsion spreads the cloud of droplets sideways. This is part of the reason why the spray widens after the nozzle.

Also accepted

  • In part a the strength of the field may be given in the unit $\text{kV/m}$: $300\,\text{kV/m}$.
  • Part b may also be solved through the acceleration: $a = \dfrac{F}{m} = 450\,\text{m/s}^2$ and $v = \sqrt{2 a d}$, as long as the solution is based on the force of part a and the distance $0.20\,\text{m}$.

Common mistakes

  • The potential difference $60\,\text{kV}$ is substituted in kilovolts, so the field comes out wrong by a factor of a thousand.
  • The force is calculated as $U : d$, which mixes up the strength of the field and the force.
  • In part b the formula $v = \sqrt{2 a d}$ is used with the acceleration $g$, or the work is calculated from the weight.
  • The energy of the droplet is said to "turn into heat" or to "disappear"; the energy is transferred to the air and from there as heat.
  • In part d the distance is substituted in millimetres, or the squaring is forgotten.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)4 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. In a uniform field $E = \dfrac{U}{d}$ and the force $F = qE$ acts on the droplet. 1 p
The numerical values are substituted with their units, and the calculation is correct. $E = \dfrac{60\,000\,\text{V}}{0.20\,\text{m}} = 3.0 \cdot 10^{5}\,\text{V/m}$ 1 p
The answer has the correct unit and a precision that matches the given data. $F \approx 1.8 \cdot 10^{-7}\,\text{N}$ 1 p
The reasonableness of the result has been assessed. The weight $mg \approx 3.9 \cdot 10^{-9}\,\text{N}$ is only about $2\,\%$ of the electric force, so it can be ignored. The field is about a tenth of the dielectric strength of air, so no sparking occurs. 1 p

b)4 p

Chooses the principle of energy: the work done by the electric field, $qU$, increases the kinetic energy of the droplet when weight and air resistance are left out; the strength of the field or the acceleration is not needed. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $q U = \dfrac{1}{2} m v^2$, so $v = \sqrt{\dfrac{2 q U}{m}}$; $U$ is the potential difference between the nozzle plate and the workpiece. 1 p
The numerical values are substituted with their units, and the calculation is correct. $qU = 6.0 \cdot 10^{-13}\,\text{C} \cdot 6.0 \cdot 10^{4}\,\text{V} = 3.6 \cdot 10^{-8}\,\text{J}$ and $v = \sqrt{\dfrac{2 \cdot 3.6 \cdot 10^{-8}\,\text{J}}{4.0 \cdot 10^{-10}\,\text{kg}}}$. 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 13\,\text{m/s}$ 1 p

c)4 p

Recognises that the model of part b left out air resistance: air resists the motion of the droplet, and as the speed increases, air resistance increases until it is approximately equal to the electric force. 1 p
Calculates the actual kinetic energy $\dfrac{1}{2} m v^2 = 5.0 \cdot 10^{-9}\,\text{J}$ and compares it with the work done by the field, $3.6 \cdot 10^{-8}\,\text{J}$: it is only about $14\,\%$. 1 p
Concludes that the remaining about $3.1 \cdot 10^{-8}\,\text{J}$ is transferred to the air: the internal energy of the air increases and the energy is transferred further as heat to the surroundings. Energy is not lost. 1 p
The conclusion answers the question and is based on the data. Concludes that the result of part b is the greatest possible speed (an upper limit) in the absence of air resistance, and that the energy is not lost but transferred to the air. 1 p

d)4 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. Coulomb's law for point charges: $F = k\dfrac{q_1 q_2}{r^2}$. 1 p
The numerical values are substituted with their units, and the calculation is correct. Conversion $1.0\,\text{mm} = 1.0 \cdot 10^{-3}\,\text{m}$; $F = 8.988 \cdot 10^{9}\,\tfrac{\text{N}\cdot\text{m}^2}{\text{C}^2} \cdot \dfrac{(6.0 \cdot 10^{-13}\,\text{C})^2}{(1.0 \cdot 10^{-3}\,\text{m})^2}$. 1 p
The answer has the correct unit and a precision that matches the given data. $F \approx 3.2 \cdot 10^{-9}\,\text{N}$ 1 p
The conclusion answers the question and is based on the data. Charges of the same sign repel each other, so the cloud of droplets spreads sideways. The force is about $2\,\%$ of the force of the field but almost as large as the weight of the droplet ($0.8$ times), so the droplets still move towards the workpiece but spread out. 1 p

Self-assessed points: 0 / 16 (16 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Camera flash 14 points

Made-up example data · The values of the capacitor are of the same order of magnitude as in a camera flash unit, but they have been chosen for the task.

In a camera flash unit a small battery charges a capacitor to a voltage of $300\,\text{V}$ with the help of a circuit that raises the voltage. The capacitance of the capacitor is $330\,\mu\text{F}$. When the flash fires, the capacitor discharges through the flash tube in about $1.0\,\text{ms}$, and its voltage falls to $220\,\text{V}$. The energy transferred to the flash leaves the device as light and heat.

  1. a) How much energy is stored in the capacitor before the flash? 3 p
  2. b) How much energy was transferred to the flash? What fraction of the stored energy is this, and why is the fraction larger than the decrease in voltage? 4 p
  3. c) How large was the average current that flowed from the capacitor during the flash? 3 p
  4. d) A repairer removes the battery from the flash unit and opens the unit right after that. Evaluate whether there is any danger in the capacitor, and describe how a capacitor is discharged safely. Also assess the danger by calculating the current if the resistance of a human body is about $1.0\,\text{k}\Omega$. 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

$C = 330\,\mu\text{F}$ and $U = 300\,\text{V}$

Find

the energy $E$ stored in the capacitor before the flash

Principle

Energy is stored in a charged capacitor, because charge has been moved onto its plates against the voltage. The energy depends on the capacitance and the square of the voltage.

Formula

$E = \dfrac{1}{2} C U^2$

Conversions

$330\,\mu\text{F} = 330 \cdot 10^{-6}\,\text{F}$

Substitution

$E$$= \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot (300\,\text{V})^2$
$= \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot 90\,000\,\text{V}^2$
$= 14.85\,\text{J}$

Answer

$E \approx 15\,\text{J}$

($14.85\,\text{J}$), two significant figures, because the precision of the voltage $300\,\text{V}$ and the capacitance $330\,\mu\text{F}$ is at most two or three figures.

Reasonableness

The energy is of the order of a joule, as in the capacitor of a small flash unit. If this energy were used to lift an object with a weight of one kilogram, it would rise about $1.5\,\text{m}$.

b)

Given

$C = 330\,\mu\text{F}$, the voltage before the flash $U_1 = 300\,\text{V}$ and after it $U_2 = 220\,\text{V}$

Find

the energy $\Delta E$ transferred to the flash, its fraction of the stored energy and the reason for the size of the fraction

Principle

The energy transferred to the flash is the change in the energy of the capacitor: it is the energy before the flash minus the energy after the flash. The voltage is squared in the energy formula, so each energy is calculated separately; the change in voltage $\Delta U$ alone is not enough.

Formula

$\Delta E$$= \dfrac{1}{2} C U_1^2 - \dfrac{1}{2} C U_2^2$

Conversions

$330\,\mu\text{F} = 330 \cdot 10^{-6}\,\text{F}$

Substitution

$E_1$$= 14.85\,\text{J}$
$E_2$$= \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot (220\,\text{V})^2$
$= \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot 48\,400\,\text{V}^2$
$= 7.99\,\text{J}$

so

$\Delta E$$= 14.85\,\text{J} - 7.99\,\text{J}$
$= 6.86\,\text{J}$

The fraction is

$\dfrac{6.86\,\text{J}}{14.85\,\text{J}}$$= 0.46$

Answer

The energy transferred to the flash was $\Delta E \approx 6.9\,\text{J}$, about $46\,\%$ of the stored energy, two significant figures as in part a.

Reasonableness

The voltage decreased by only

$\dfrac{80\,\text{V}}{300\,\text{V}}$$\approx 27\,\%$

but $46\,\%$ of the energy was transferred, because energy is proportional to the square of the voltage:

$1 - \left(\dfrac{220}{300}\right)^2$$= 0.46$

The transferred energy is smaller than the initial energy, as it should be.

c)

Given

$C = 330\,\mu\text{F}$, change in voltage $\Delta U = 300\,\text{V} - 220\,\text{V} = 80\,\text{V}$ and duration $\Delta t = 1.0\,\text{ms}$

Find

the average current $I$ during the flash

Principle

Current is the rate at which charge is transferred. The charge of the capacitor is $Q = C U$, so when the voltage decreases by $80\,\text{V}$, a charge $\Delta Q = C\,\Delta U$ has left the capacitor through the flash tube.

Formula

$I = \dfrac{\Delta Q}{\Delta t}$ and $\Delta Q = C\,\Delta U$

Conversions

$330\,\mu\text{F} = 330 \cdot 10^{-6}\,\text{F}$ and $1.0\,\text{ms} = 1.0 \cdot 10^{-3}\,\text{s}$

Substitution

$\Delta Q$$= 330 \cdot 10^{-6}\,\text{F} \cdot 80\,\text{V}$
$= 0.0264\,\text{C}$
$I$$= \dfrac{0.0264\,\text{C}}{1.0 \cdot 10^{-3}\,\text{s}}$
$= 26.4\,\text{A}$

Answer

$I \approx 26\,\text{A}$

Two significant figures like the precision of the duration $1.0\,\text{ms}$.

Reasonableness

The average power is

$\dfrac{6.86\,\text{J}}{1.0 \cdot 10^{-3}\,\text{s}}$$\approx 6.9\,\text{kW}$

and the current $26\,\text{A}$: these are much larger than a small battery could supply in such a short pulse. That is why the energy is first stored in the capacitor, which releases it quickly.

d)

Find

whether there is any danger in the capacitor after the battery is removed, how it is discharged safely and how large a current would flow through the body

Principle

A capacitor stores charge and energy on its plates, and its voltage does not disappear when the battery that charges it is removed. The danger depends on the stored energy and the voltage, not on the battery. The size of the current through the body is obtained from the voltage and the resistance.

Model answer

Danger. After the battery is removed, the capacitor is still at $300\,\text{V}$ and holds about $15\,\text{J}$. The device does not get its voltage from the small battery but from the capacitor, which can discharge through the body of the repairer.

Current. If the resistance of the body is $1.0\,\text{k}\Omega$, the initial current would be $I = \dfrac{U}{R} = \dfrac{300\,\text{V}}{1.0 \cdot 10^{3}\,\Omega} = 0.30\,\text{A}$. This is many times the few tens of milliamperes that can already be dangerous. The current decreases as the capacitor discharges, but the initial current is enough for a dangerous electric shock.

Safe discharge (only for understanding: do not try to discharge a capacitor yourself). A suitable resistor, for example $10\,\text{k}\Omega$, is connected between the terminals of the capacitor with insulated tools. The initial current is then $\dfrac{300\,\text{V}}{10\,\text{k}\Omega} = 30\,\text{mA}$, and the energy of the capacitor is transferred to the internal energy of the resistor and from there as heat to the surroundings. At first the power of the resistor is $\dfrac{(300\,\text{V})^2}{10\,\text{k}\Omega} = 9\,\text{W}$, so the power rating of the resistor must be suitable. The terminals are not short-circuited with a screwdriver, because the current is then very large and a spark is produced. Finally the voltage is checked with a meter.

Conclusion. The device must not be opened right away. The danger comes from the energy and voltage stored in the capacitor, not from the size of the battery. The resistance of the body varies, for example with the moisture of the skin, so the calculated current is an order-of-magnitude estimate.

Also accepted

  • In part a the energy may also be calculated with the formula $E = \dfrac{1}{2} Q U$, where $Q = C U = 0.099\,\text{C}$.
  • In part b the fraction may be calculated directly from the ratio of the voltages, $1 - \left(\dfrac{U_2}{U_1}\right)^2$.
  • In part c the result may also be calculated as the ratio of charge and time without a separate calculation of $\Delta Q$: $I = \dfrac{C\,\Delta U}{\Delta t}$.

Common mistakes

  • The capacitance is substituted in microfarads without multiplying it by $10^{-6}$.
  • In calculating the energy the squaring of the voltage is forgotten, or the energy transferred to the flash is calculated with the formula $\tfrac{1}{2} C (\Delta U)^2$.
  • The current is calculated by dividing the energy by the time, which gives the power.
  • It is stated that the capacitor is harmless because the device has only a small battery.
  • The capacitor is discharged by short-circuiting it with a metal tool.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. The energy of a capacitor $E = \dfrac{1}{2} C U^2$. 1 p
The numerical values are substituted with their units, and the calculation is correct. Conversion $330\,\mu\text{F} = 330 \cdot 10^{-6}\,\text{F}$; $E = \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot (300\,\text{V})^2$. 1 p
The answer has the correct unit and a precision that matches the given data. $E \approx 15\,\text{J}$ 1 p

b)4 p

Concludes that the energy transferred to the flash is the change in the energy of the capacitor: the energies are calculated separately before and after, not from the change in voltage $\Delta U$ alone. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\Delta E = \dfrac{1}{2} C U_1^2 - \dfrac{1}{2} C U_2^2$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $E_2 = \dfrac{1}{2} \cdot 330 \cdot 10^{-6}\,\text{F} \cdot (220\,\text{V})^2 = 7.99\,\text{J}$ and $\Delta E = 14.85\,\text{J} - 7.99\,\text{J} = 6.9\,\text{J}$. 1 p
The reasonableness of the result has been assessed. The fraction is about $46\,\%$, although the voltage decreased by only about $27\,\%$, because energy depends on the square of the voltage: $1 - \left(\dfrac{220}{300}\right)^2 \approx 0.46$. 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $I = \dfrac{\Delta Q}{\Delta t}$, where the transferred charge is $\Delta Q = C\,\Delta U$. 1 p
The numerical values are substituted with their units, and the calculation is correct. $\Delta Q = 330 \cdot 10^{-6}\,\text{F} \cdot 80\,\text{V} = 0.0264\,\text{C}$ and $\Delta t = 1.0 \cdot 10^{-3}\,\text{s}$. 1 p
The reasonableness of the result has been assessed. $I \approx 26\,\text{A}$ and the average power $\dfrac{6.9\,\text{J}}{1.0\,\text{ms}} \approx 6.9\,\text{kW}$ are much larger than a small battery could supply in such a short pulse; that is why the energy is first stored in the capacitor. 1 p

d)4 p

Knows that the capacitor keeps its charge and energy when the battery is removed: it is still at $300\,\text{V}$ and holds about $15\,\text{J}$, even though the device is supplied by a small battery. 1 p
Calculates the current through the body: $I = \dfrac{U}{R} = \dfrac{300\,\text{V}}{1.0 \cdot 10^{3}\,\Omega} = 0.30\,\text{A}$, which is many times the few tens of milliamperes that can already be dangerous. 1 p
Describes a safe discharge: a suitable resistor, for example $10\,\text{k}\Omega$, is connected between the terminals with insulated tools, so that the energy is transferred to the internal energy of the resistor and further as heat to the surroundings; the initial current is $\dfrac{300\,\text{V}}{10\,\text{k}\Omega} = 30\,\text{mA}$. The terminals are not short-circuited, and the voltage is checked with a meter. At the start the resistor must withstand about $\dfrac{(300\,\text{V})^2}{10\,\text{k}\Omega} = 9\,\text{W}$, so its power rating must be chosen accordingly. The description is meant for understanding: do not try to discharge it yourself. 1 p
The conclusion answers the question and is based on the data. Concludes that the device must not be opened right away: the danger comes from the stored energy and voltage, not from the size of the battery. The resistance of the body varies (for example, moist skin), so the estimate is an order of magnitude. 1 p

Self-assessed points: 0 / 14 (14 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).