2023 A/L Physics Past Paper · Question 39 (English Medium)

Answer: (5) $6\,\text{V}$, $18\,\mu\text{C}$, $54\,\mu\text{J}$

Given information

What we know:

The two capacitors are $C_1 = 3\,\mu\text{F}$ and $C_2 = 6\,\mu\text{F}$.

They are in series with a battery of $V = 9\,\text{V}$.

The circuit has reached steady state.

What the question asks

We want: the voltage, the charge and the energy of the $3\,\mu\text{F}$ capacitor.

What each letter means:

$C$ ($\mu\text{F}$): the equivalent capacitance of the two capacitors.

$Q$ ($\mu\text{C}$): the charge on each capacitor.

$V_1$ ($\text{V}$): the voltage across the $3\,\mu\text{F}$ capacitor.

$E$ ($\mu\text{J}$): the energy stored in the $3\,\mu\text{F}$ capacitor.

Relevant theory

If a number of capacitors are connected so that they carry the same charge then, these capacitors are said to be in series.

This clearly shows that a charged capacitor contains energy and this energy gets converted to other forms such as heat during the discharge.

The first quote says that capacitors in series carry the same charge. The second quote says that a charged capacitor holds energy.

For example, think of two buckets joined by a pipe, with a pump at one end. The same amount of water passes through the pipe into both buckets.

A capacitor is like a bucket. The wide one fills with a little push, and the narrow one needs a big push. So the small capacitor takes the bigger share of the voltage.

Required equations

$C = \dfrac{C_1 C_2}{C_1 + C_2}$

$Q = C\,V$

$V_1 = \dfrac{Q}{C_1}$

$E = \dfrac{1}{2}C_1 V_1^2$

Here $\mu\text{F}$ times $\text{V}$ gives $\mu\text{C}$, and $\dfrac{1}{2} \times \mu\text{F} \times \text{V}^2$ gives $\mu\text{J}$.

Applying it to this question

$2$

$C = \dfrac{3 \times 6}{3 + 6}$

$18$

$Q = C\,V = 2 \times 9$

$6$

$V_1 = \dfrac{Q}{C_1} = \dfrac{18}{3}$

$54$

$E = \dfrac{1}{2} \times 3 \times 6^2$

$c1 * c2 / (c1 + c2)$

$0.5 * c * pow(v, 2)$

Reasoning and calculation

Now match it with the five options.

(1) $3\,\text{V}$, $9\,\mu\text{C}$, $27\,\mu\text{J}$ does not match.

(2) $3\,\text{V}$, $9\,\mu\text{C}$, $54\,\mu\text{J}$ does not match.

(3) $3\,\text{V}$, $18\,\mu\text{C}$, $108\,\mu\text{J}$ does not match.

(4) $6\,\text{V}$, $18\,\mu\text{C}$, $27\,\mu\text{J}$ does not match, because the energy is $54\,\mu\text{J}$.

(5) $6\,\text{V}$, $18\,\mu\text{C}$, $54\,\mu\text{J}$ matches.

Checks

Unit check: $Q$ is in $\mu\text{C}$, $V_1$ in $\text{V}$ and $E$ in $\mu\text{J}$. Each value comes out in the unit we asked for.

Does it make sense?

The $6\,\mu\text{F}$ capacitor has $\dfrac{18}{6} = 3\,\text{V}$ across it. Add the two voltages: $6 + 3 = 9\,\text{V}$. That is the battery voltage, as it must be.

The smaller capacitor, $3\,\mu\text{F}$, takes the bigger voltage, $6\,\text{V}$.

Conclusion

$V_1 = 6\,\text{V},\ Q = 18\,\mu\text{C},\ E = 54\,\mu\text{J}$

The $3\,\mu\text{F}$ capacitor has $6\,\text{V}$ across it, holds $18\,\mu\text{C}$ and stores $54\,\mu\text{J}$.

The answer is (5).

The quotes give the same charge on capacitors in series and the energy stored in a charged capacitor. The equivalent value $\dfrac{C_1 C_2}{C_1 + C_2}$ and $E = \dfrac{1}{2}CV^2$ follow from adding the voltages across the two capacitors and from the quoted rules.

Teacher's note

In series the charge is the same on each capacitor. Find it from the equivalent capacitance, $Q = CV$, then use $V_1 = \dfrac{Q}{C_1}$ and $E = \dfrac{1}{2}C_1V_1^2$.

The voltages must add up to the battery voltage.

Exam tip