Question 2.54: In the circuit shown in Fig. 2.166 the capacitor is fully ch......
In the circuit shown in Fig. 2.166 the capacitor is fully charged when the switch is closed. Calculate the voltage across the fully charged capacitor. Also calculate the voltage across the capacitor and the current in the capacitor circuit 0.05 seconds after opening of the switch.
When the capacitor is fully charged, no current will flow through the capacitor. The capacitor will be charged to a voltage equal to V_{PQ}. However, current, I flowing through the loop PLMQP will be
I=\frac{12}{(2+4+6) \times 10^3} \mathrm{~A}Voltage across terminals P and Q, i.e., V_{PQ} is
V_{PQ} = IR drops across 4 kΩ and 6 kΩ resistors
Now, when the switch is opened, at t = 0, v_c = 10 V. The capacitor will be getting discharged through the resistors 10 kΩ, 4 kΩ, and 6 kΩ in the loop RPLMQSR. The time constant of the circuit, τ = RC.
\begin{aligned} \tau & =R C=(10+4+6) \times 10^3 \times 2.5 \times 10^{-6}=20 \times 10^3 \times 2.5 \times 10^{-6} \\ & =0.05 \text { seconds } \end{aligned}Let the capacitor voltage at t = 0.05 sec be v_c^{\prime} \text { and discharging current at } t=0.05 \text { be } i^{\prime} .
v_c^{\prime}=v_c e^{-t / \tau}=10 e^{\frac{-0.05}{0.05}}=10 \times e^{-1}=10 \times 0.368=3.68 \mathrm{~V}Initial current at t = 0, I_0=\frac{V_{A B}}{(4+6) \times 10^3}=\frac{10}{10 \times 10^3}=1 \mathrm{~mA}
Current after 0.05 sec, i^{\prime}=I_0 e^{-1}=1 \times 0.368 \mathrm{~mA}=0.368 \mathrm{~mA}