Encyclopedia of Electrical Engineering

Find the voltage across and charge on capacitor C1 of the network shown in the figure below after it has charged up to its final value.

2021-07-24 15:44:03

1 Answer

As we already know that capacitor is effectively an open circuit for dc sources after charging up to its final value. The network can be redrawn as by removing capacitor from its branch assuming capacitor act as an open circuit.

Therefore,

$$\begin{split}V_C &= {R_2 \, E \over R_2 + R_1}\\&= {8 \times 24 \over 8 + 4}\\&= 16 V \end{split}$$

$$\begin{split} Q_1 &= C_1 \, V_C\\&= (20\times 10^{-6})(16V)\\&= 320 \mu F\\\end{split}$$

2021-07-25 08:48:47