2020 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (1) $0.5\ \text{V}$

Given information

Given: left junction $L$, right junction $R$, three branches between them — top: a $2\ \text{V}$ cell and $3\ \Omega$; middle: a $3\ \text{V}$ cell and the ideal capacitor $C$; bottom: a $6\ \text{V}$ cell and $5\ \Omega$.

Every cell's positive terminal (long line) faces left; internal resistances are negligible.

What the question asks

We must find: the steady-state potential difference across capacitor $C$.

Relevant theory

That is, each of these capacitors has an electric insulator pressed between two electric conductors.

The reason why the microammeter now indicates zero is that when the capacitor is charged to the maximum, the circuit gets disconnected.

Also, if the internal resistance (r) of the source is zero V=E.

"The algebraic sum of the currents flowing across a junction of a circuit is zero".

"The algebraic sum of the electromotive forces of a closed circuit is equal to the algebraic sum, of the products of the currents flowing along all branches of the circuit and their respective resistances".

Required equations

$I_C = 0$

$\Sigma E = \Sigma IR$

$V_L - V_R = E_1 + IR_1$

$V_L - V_R = E_2 + V_C$

Applying it to this question

Plan: (1) no current flows through the charged capacitor, so by Kirchhoff's first law at junction $L$, the top branch's current returns through the bottom branch — the outer loop carries one current $I$.

(2) find it from Kirchhoff's second law; (3) find $V_L - V_R$; (4) find $V_C$ from the middle branch.

Reasoning and calculation

$\Sigma E = 6 - 2 = 4\ \text{V}$

$\Sigma R = 3 + 5 = 8\ \Omega$

$0.5$

$I = \frac{4}{8}$

$3.5$

$V_L - V_R = 2 + 0.5 \times 3$

$V_L - V_R = 6 - 0.5 \times 5$

$3.5 = 3 + V_C$

$V_C = 3.5 - 3$

Comparing with the options: $V_C = 0.5\ \text{V}$ — option (1); the capacitor's left plate (connected to the middle cell's negative terminal) is at the higher potential by $0.5\ \text{V}$.

Checks

$E1 + R1 * (E3 - E1) / (R1 + R3) - E2$

Second way: compute $V_L - V_R$ via the top branch ($2 + 1.5$) and the bottom branch ($6 - 2.5$) — both give $3.5\ \text{V}$ — the p.d. between the same two junctions must agree ✓.

Power check: the $6\ \text{V}$ cell delivers $6 \times 0.5 = 3\ \text{W}$; the $2\ \text{V}$ cell absorbs $1\ \text{W}$, the resistors dissipate $0.5^2 \times 8 = 2\ \text{W}$ — $3 = 1 + 2$ ✓.

Physical plausibility: $V_C$ is smaller than the junctions' $3.5\ \text{V}$ because most of it belongs to the middle cell; if the middle cell were $3.5\ \text{V}$, $V_C = 0$ — the capacitor would not charge at all ✓.

Unit check: $\text{A} \times \Omega = \text{V}$; every step only adds volts ✓.

Conclusion

$V_C = 0.5\ \text{V}$

No current flows in the middle branch; the outer loop carries $I = 0.5\ \text{A}$, $V_L - V_R = 3.5\ \text{V}$; $V_C = 3.5 - 3 = 0.5\ \text{V}$ — the answer is (1).

Taking the p.d. between the junctions as the p.d. across the capacitor.

The middle branch has a cell too: $V_L - V_R = E_2 + V_C$.