The charge accumulated on the capacitor connected in the following circuit is $______$ $\mu C$. (Given $C=150$ $\mu F$)

Question

Aicharya · Accepted Answer

Under the balanced condition, the capacitor will act as an open circuit.

The current $(i_{1})$ through the $1 Ω$ resistor is given by

$\begin{array}{rcl} i_{1} & = & \frac{10 V}{1 Ω + 2 Ω} \\ = & \frac{10}{3} A \end{array}$

The current $(i_{2})$ through the $6 Ω$ resistor is given by

$\begin{array}{rcl} i_{2} & = & \frac{10 V}{6 Ω + 4 Ω} \\ = & 1 A \end{array}$

From the above diagram, it can be written that

$V_{A} + i_{1} \times 1 - i_{2} \times 6 = V_{B} \Rightarrow V_{A} - V_{B} = 1 \times 6 - \frac{10}{3} \times 1 = \frac{8}{3} V$

Hence, the charge on the capacitor is given by

$\begin{array}{rcl} Q & = & C (V_{A} - V_{B}) \\ = & 150 μF \times \frac{8}{3} V \\ = & 400 μC \end{array}$

The charge accumulated on the capacitor connected in the following circuit is $______$ $\mu C$. (Given…