In Fig. key K is closed at t=0. After a long time

Sahay Sir > Question Answers > In Fig. key K is closed at t=0. After a long time
An inductor having inductance L and a capacitor having capacitance C are connected in series. The current in the circuit increases linearly in time as described by I = Kt. The capacitor is initially uncharged. (To answer the following, use L, C, K, and t as necessary.) a) Determine the voltage across the inductor as a function of time. b)Determine the voltage across the capacitor as a function of time. c)Determine the time when the energy stored in the capacitor first exceeds that in the inductor.
22
Oct
An inductor having inductance L and a capacitor having capacitance C are connected in series. The current in the circuit increases linearly in time as described by I = Kt. The capacitor is initially uncharged. (To answer the following, use L, C, K, and t as necessary.) a) Determine the voltage across the inductor as [...]
Posted in: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Tags: C = 2μF , E 1 = 3 V and E 2 = 7 V , In Fig. key K is closed at t=0. After a long time , L = 4 mH , respectively , the potential difference between A and B is zero , the value of R will be [ r 1 = r 2 = 1 Ω , where r1 and r2 are the internal resistances of cells E1 and E2 ,
The cell in the circuit shows in Fig is ideal. The coil has an inductance of 4 mH and a resistance of 2mΩ. The switch is closed at t=0. The amount of energy stored in the inductor at t = 2 s is (take e = 3)
22
Oct
The cell in the circuit shows in Fig is ideal. The coil has an inductance of 4 mH and a resistance of 2mΩ. The switch is closed at t=0. The amount of energy stored in the inductor at t = 2 s is (take e = 3) C = 2μF E 1 = 3 V [...]
Posted in: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Tags: C = 2μF , E 1 = 3 V and E 2 = 7 V , In Fig. key K is closed at t=0. After a long time , L = 4 mH , respectively , the potential difference between A and B is zero , the value of R will be [ r 1 = r 2 = 1 Ω , where r1 and r2 are the internal resistances of cells E1 and E2 ,
In Fig. key K is closed at t=0. After a long time, the potential difference between A and B is zero, the value of R will be [ r 1 = r 2 = 1 Ω, E 1 = 3 V and E 2 = 7 V, C = 2μF, L = 4 mH, where r1 and r2 are the internal resistances of cells E1 and E2, respectively].
22
Oct
In Fig. key K is closed at t=0. After a long time, the potential difference between A and B is zero, the value of R will be [ r 1 = r 2 = 1 Ω, E 1 = 3 V and E 2 = 7 V, C = 2μF, L = 4 mH, where [...]
Posted in: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Tags: C = 2μF , E 1 = 3 V and E 2 = 7 V , In Fig. key K is closed at t=0. After a long time , L = 4 mH , respectively , the potential difference between A and B is zero , the value of R will be [ r 1 = r 2 = 1 Ω , where r1 and r2 are the internal resistances of cells E1 and E2 ,