Question Answers

Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > Initially the 900μF capacitor is charged to 100 V and the 100μF capacitor is uncharged in Fig. Then switch S2 is closed for time t1, after which it is opened and at the same instant switch S1 is closed for time t2 and then opened. It is now found that 100μF capacitor is charged to 300μF capacitor is charged to 300 V. Find the minimum possible value of the time interval t1 and t2.

Initially the 900μF capacitor is charged to 100 V and the 100μF capacitor is uncharged in Fig. Then switch S2 is closed for time t1, after which it is opened and at the same instant switch S1 is closed for time t2 and then opened. It is now found that 100μF capacitor is charged to 300μF capacitor is charged to 300 V. Find the minimum possible value of the time interval t1 and t2.

a or b c or d E = 120 V In the circuit shown in fig In the circuit shown in Figure Initially the 900μF capacitor is charged to 100 V and the 100μF capacitor is uncharged in Fig. Then switch S2 is closed for time t1 is at higher potential? (c) What is the potential difference Vcd across the inductor L? (d) Which point is at steady state. if R1 = R2= R3 =R and the current through each resistor is i. Find the currents through the resistors immediately after the switch S is opened. R1 = 30.0 Ω R2 =50.0Ω and L = 0.200 H. Switch S is closed at t=0. Just after the switch is closed.(a) What is the potential difference Vab across the inductor R1? (b) Which point