Five capacitors of capacitances C1 = C2 = C3 = C4 = 10 μF and C5 = 2.5 μF are connected as shown, along with a battery of 50 V. [figure] The equivalent capacitance and the charges on each capacitor respectively are :
- 4 μF, 250 μC on C1 to C4 and 125 μC on C5
- 5 μF, 250 μC on all capacitors
- 5 μF, 125 μC on C1 to C4 and 25 μC on C5
- 5 μF, 125 μC on all capacitors
Answer
(4) 5 μF, 125 μC on all capacitors
Solution
C1, C2, C3, C4 form one series branch: 1C' = 410 ⇒ C' = 2.5 μF
This branch and C5 are both connected directly across the 50 V battery ⇒ parallel
∴ Ceq = C' + C5 = 2.5 + 2.5 = 5 μF
Series branch: same charge on each, q = C'V = 2.5 50 = 125 μC on each of C1 to C4
C5: q5 = C5V = 2.5 50 = 125 μC
∴ 5 μF and 125 μC on all capacitors
Answer: (4)