In the circuit shown in figure `R_1=30Omega, R_2=40 Omega, L=0.4H` and `C=1/3mF`. Find seven fun...

For the ciruit shown in figure E=50V , R_1=10Omega, R_2=20Omega, R_3=30Omega and L=2.0 mH . Find a. Immediately after switch S is closed b. A long time after S is closed. c. Immediaately after S is reopened d. A long time after S is reopened.

In the circuit shown in figure E_1=10 V, E_2=4V, r_1=r_2=1Omega and R=2Omega . Find the potential difference across battery 1 and battery 2.

In the circuit shown in figure E_1=7V,E_2=1 V,R_1=2Omega, R_2=2Omega and R_3=3Omega respectively. Find the power supplied by the two batteries.

In the circuit shown in figure, R(1) = 1Omega, R_(2) = 2Omega, C_(1) = 1 muF, C_(2) = 2 muF and E = 6 V . Calculate charge on each capacitor in steady state

In the circuit shown in fig. R_(1) = 10 Omega, R_(2) = Omega, C_(1) = 1 muF, C_(2) = 2muF, and E =6V. Calculate the charge on each capactor in the stedy state.

In the circuit shown in the figure R_(1) =3 Omega,R_(2) = 2 Omega and R_(3) = 5 Omega Emf of the cell epsilon = 6 Volt. (a) Infinitely many 1 Omega resistors are added in parallel to R_(1) and R_(2) . Find current through R_(2) and potential drop across it. (b) Infinitely many 1 Omega resistors are added in series to R_(3) . Find the potential drop across R_(3) and current through it.

In the circuit shown in fig. X_(C ) = 100 Omega,(X_L)=200 Omega and R=100 Omega . The effective current through the source is

In the circuit shown in figure : R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega and C = (sqrt(3))/(2) mF . Current in L - R_(1) circuit is I_(1) in C - R_(1) circuit is I_(2) and the main current is I At some instant current in L - R_(1) circuit is 10 A . At the same instant current in C - R_(2) branch will be

In the circuit shown in figure : R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega and C = (sqrt(3))/(2) mF . Current in L - R_(1) circuit is I_(1) in C - R_(1) circuit is I_(2) and the main current is I At some instant I_(1) in the circuit is 10 sqrt(2) A , then at this instant current I will be

In the circuit shown in figure : R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega and C = (sqrt(3))/(2) mF . Current in L - R_(1) circuit is I_(1) in C - R_(1) circuit is I_(2) and the main current is I Phase difference between I_(1) and I_(2) is