First slide

Capacitance

Question

A capacitor of capacitance C₁ = 1 $μ$ F charged up to a voltage V = 110 V is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances C₂ = 2 $μ$ F and C₃ = 3 $μ$ F. Then, the amount of charge that will flow through the connecting wires is

Moderate

A

40 $μ$ C
B

50 $μ$ C
C

60 $μ$ C
D

30 $μ$ C

Solution

$Q_{1} = C_{1} V_{1} = 110 μC$
$Q_{1} = 10^{- 6} (1 + \frac{6}{5}) V = 2.2 V x 10^{- 6}$
110 $μ$ C = V x 2.2 x10^-6
$∴$ V = 50 V
Q = C₁V = 50 $μ$ C
charge flow in the wire
= Q₁ - Q = (110 - 50) $μ$ C = 60 $μ$ C

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App