Question

Each capacitor in the combination shown in figure has a breakdown voltage of 20 V. What is the breakdown voltage (in V), of the combination?

Answer 1

30

Answer 2

To determine the breakdown voltage of the combination, we need to analyze the circuit step-by-step, considering both parallel and series connections.

1. Analyze the parallel combinations: The circuit consists of two parallel blocks and one single capacitor connected in series.

• First parallel block (left side): Two 20 µF capacitors are connected in parallel.

  • Equivalent capacitance: $C_{P1} = 20 \mu F + 20 \mu F = 40 \mu F$.
  • Breakdown voltage: Since capacitors in parallel have the same voltage across them, the breakdown voltage of the parallel combination is limited by the individual breakdown voltage of any capacitor in that block. As each capacitor has a breakdown voltage of 20 V, the breakdown voltage for this block is $V_{P1, max} = 20 V$.
  • Maximum charge this block can hold: $Q_{P1, max} = C_{P1} \times V_{P1, max} = 40 \mu F \times 20 V = 800 \mu C$.

• Second parallel block (right side): Two 20 µF capacitors are connected in parallel.

  • Equivalent capacitance: $C_{P2} = 20 \mu F + 20 \mu F = 40 \mu F$.
  • Breakdown voltage: Similarly, $V_{P2, max} = 20 V$.
  • Maximum charge this block can hold: $Q_{P2, max} = C_{P2} \times V_{P2, max} = 40 \mu F \times 20 V = 800 \mu C$.

• Middle capacitor: This is a single 10 µF capacitor.

  • Capacitance: $C_S = 10 \mu F$.
  • Breakdown voltage: $V_{S, max} = 20 V$.
  • Maximum charge this capacitor can hold: $Q_{S, max} = C_S \times V_{S, max} = 10 \mu F \times 20 V = 200 \mu C$.

2. Analyze the series combination: Now, we have three equivalent capacitors in series: $C_{P1}$ (40 µF), $C_S$ (10 µF), and $C_{P2}$ (40 µF). In a series combination, the charge stored on each capacitor is the same. The overall breakdown voltage of the series combination is reached when the voltage across any one of the series components reaches its individual breakdown voltage. This happens when the total charge on the series combination reaches the minimum of the maximum charges that each individual component can safely hold.

• Maximum charge for the series combination: $Q_{total, max} = \min(Q_{P1, max}, Q_{S, max}, Q_{P2, max})$ $Q_{total, max} = \min(800 \mu C, 200 \mu C, 800 \mu C)$ $Q_{total, max} = 200 \mu C$.

• When this maximum charge (200 µC) is applied to the series combination, we calculate the voltage across each component:

  • Voltage across $C_{P1}$: $V_{P1} = \frac{Q_{total, max}}{C_{P1}} = \frac{200 \mu C}{40 \mu F} = 5 V$.
  • Voltage across $C_S$: $V_S = \frac{Q_{total, max}}{C_S} = \frac{200 \mu C}{10 \mu F} = 20 V$.
  • Voltage across $C_{P2}$: $V_{P2} = \frac{Q_{total, max}}{C_{P2}} = \frac{200 \mu C}{40 \mu F} = 5 V$.

• At this point, the 10 µF capacitor ($C_S$) reaches its breakdown voltage of 20 V. The other two equivalent capacitors ($C_{P1}$ and $C_{P2}$) are only at 5 V, which is well below their breakdown limit of 20 V.

• The total breakdown voltage of the combination is the sum of the voltages across the series components at the point of breakdown: $V_{breakdown, total} = V_{P1} + V_S + V_{P2} = 5 V + 20 V + 5 V = 30 V$.

Thus, the combination will break down when the applied voltage reaches 30 V, because at this voltage, the 10 µF capacitor will fail.