2019 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (3) Circuit (3)

What the question asks

We want:

Which of the five circuits stores the most energy.

All five use the same three capacitors $C$ and the same two cells $E$.

Relevant theory

The energy in a capacitor is:

$$U = \tfrac{1}{2}CV^2$$

Capacitors in series: $\dfrac{1}{C_{eq}} = \sum \dfrac{1}{C}$. Three equal ones give $\dfrac{C}{3}$.

Capacitors in parallel: $C_{eq} = \sum C$. Three equal ones give $3C$.

Cells in series give $2E$. Cells in parallel give $E$.

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

Required equations

$U = \frac{1}{2}C_{eq}V^2$

$C_{eq} = \frac{C}{3}$

$C_{eq} = 3C$

Reading the options

The five options are the five circuits.

For each circuit find $C_{eq}$ and $V$.

Write $U$ as a multiple of $CE^2$ and compare.

Reasoning and calculation

What stays the same:

The same three capacitors $C$.

The same two cells $E$.

Every capacitor is fully charged.

What changes:

How the capacitors are joined. That changes $C_{eq}$.

How the cells are joined. That changes $V$.

$U$ grows with $C_{eq}$ and with $V^2$.

Circuits (1) and (2)

$U_1 = 0.67CE^2,\ U_2 = 0.17CE^2$

(1): capacitors in series and cells in series. $U = \tfrac{1}{2} \cdot \dfrac{C}{3} \cdot (2E)^2 = \tfrac{2}{3}CE^2$.

(2): capacitors in series and cells in parallel. $U = \tfrac{1}{2} \cdot \dfrac{C}{3} \cdot E^2 = \tfrac{1}{6}CE^2$.

Circuit (3)

$U_3 = 6CE^2$

The three capacitors are in parallel, so $3C$. The cells are in series, so $2E$.

$U = \tfrac{1}{2} \cdot 3C \cdot (2E)^2 = 6CE^2$.

Circuits (4) and (5)

$U_4 = 1.5CE^2,\ U_5 = 3CE^2$

(4): capacitors in parallel and cells in parallel. $U = \tfrac{1}{2} \cdot 3C \cdot E^2 = 1.5CE^2$.

(5): two capacitors in series give $\dfrac{C}{2}$. With the third in parallel that is $\dfrac{3C}{2}$. The cells are in series. $U = \tfrac{1}{2} \cdot \dfrac{3C}{2} \cdot (2E)^2 = 3CE^2$.

Compare in pairs.

In (1), (3) and (5) the voltage is the same, $2E$. The bigger $C_{eq}$ wins: $3C > \dfrac{3C}{2} > \dfrac{C}{3}$.

In (3) and (4) the $C_{eq}$ is the same, $3C$. The voltage $2E$ gives four times the energy.

U_3

U_5

U_4

U_1

U_2

$6CE^2 > 3CE^2 > 1.5CE^2 > 0.67CE^2 > 0.17CE^2$.

Circuit (3) has the biggest capacitance and the biggest voltage. So it stores the most.

Checks

$432$

$U_3 = \frac{1}{2} \times 6 \times 12^2$

$216$

$U_5 = \frac{1}{2} \times 3 \times 12^2$

$48$

$U_1 = \frac{1}{2} \times \frac{2}{3} \times 12^2$

Conclusion

$U_{max} = \frac{1}{2}(3C)(2E)^2 = 6CE^2$

Three capacitors in parallel give $3C$. Two cells in series give $2E$.

$$U = \tfrac{1}{2}(3C)(2E)^2 = 6CE^2$$

So the answer is option (3).

Thinking that capacitance adds in series, as resistance does.

For capacitors it is the other way round.

In series $C_{eq}$ gets smaller. In parallel it gets bigger.