Question 1.SGPYQ.8: In the circuit shown in the figure given below, the switch S......
In the circuit shown in the figure given below, the switch S is closed at time t = 0. The voltage across the inductance at t=0^{+} , is
(a) 2 V (b) 4 V
(c) -6 V (d) 8 V
Step-by-Step
At t < 0; V_C\left(0^{-}\right)=0 ; \quad i_L\left(0^{-}\right)=0
At t = 0; V_C(0)=0 ; \quad i_L(0)=0 .
At t > 0; V_C\left(0^{+}\right)=0 ; \quad i_L\left(0^{+}\right)=0
Therefore, to find voltage across inductor,
V_{ L }\left(0^{+}\right)=\frac{10}{3+\left\lgroup \frac{4 \times 4}{4+4} \right\rgroup} \times\left\lgroup \frac{4 \times 4}{4+4} \right\rgroup=\frac{10}{3+2} \times 2=4 V
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