Questions: Three capacitors are arranged as shown in the figure. The voltage Vab=1,500 V across the combination. C1 has a capacitance of 9.00 pF, C2 has a capacitance of 18.0 pF and C3 has a capacitance of 27.0 pF . What is the voltage across C2 ? a) 257 V b) 430 V c) 470 V d) 1,500 V e) 826 V

Solution Steps

The capacitors C2 and C3 are in parallel, and their combination is in series with C1.

For capacitors in parallel, the equivalent capacitance \( C_{23} \) is given by: \[ C_{23} = C_2 + C_3 \] \[ C_{23} = 18.0 \, \text{pF} + 27.0 \, \text{pF} = 45.0 \, \text{pF} \]

For capacitors in series, the total capacitance \( C_{total} \) is given by: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_{23}} \] \[ \frac{1}{C_{total}} = \frac{1}{9.00 \, \text{pF}} + \frac{1}{45.0 \, \text{pF}} \] \[ \frac{1}{C_{total}} = \frac{1}{9} + \frac{1}{45} \] \[ \frac{1}{C_{total}} = \frac{5}{45} + \frac{1}{45} = \frac{6}{45} = \frac{2}{15} \] \[ C_{total} = \frac{15}{2} = 7.5 \, \text{pF} \]

The voltage across capacitors in series is divided inversely proportional to their capacitances. The total voltage \( V_{ab} \) is 1500 V. \[ V_{C1} = V_{ab} \times \frac{C_{23}}{C_{total}} \] \[ V_{C1} = 1500 \, \text{V} \times \frac{45.0 \, \text{pF}}{7.5 \, \text{pF}} \] \[ V_{C1} = 1500 \, \text{V} \times 6 = 9000 \, \text{V} \]

Since C2 and C3 are in parallel, they share the same voltage. The voltage across the series combination of C2 and C3 is: \[ V_{C2} = V_{C3} = V_{ab} - V_{C1} \] \[ V_{C2} = 1500 \, \text{V} - 9000 \, \text{V} = -7500 \, \text{V} \]

Final Answer