In the circuit it is given that `R_(1)=R_(2)=R_(3)=200Omega` ,the reading of voltmetre =`100V` ,then find the e.m.f of the battery in

In the circuit diagram show, X_(C)=100Omega, X_(L)=200Omega,& R=100Omega

In the circuit R_(1)=400 ohm and R_(2)=200 ohm. If the resistance of the voltmeter R = 10,000 ohm, voltmeter reading is

In the given circuit the internal resistance of the 18 V cell is negligible. If R_(1)=400 Omega and the reading of an ideal voltmeter across R_(4) is 5V, then the value of R_(2) will be :

In the circuit as shown, it is given that R_1//R_2 = 1//2 and potential at O is V. The resistance R_2 is-

The power dissipated in resistor R_(3) shown in the figure is 15 W. The reading the ammeter is 500 mA and the reading of the voltmeter is 10 V. Ammeter, voltmeter and battery are ideal. Find the resistance R_(1)

The power dissipated in resistor R_(3) shown in the figure is 15 W. The reading the ammeter is 500 mA and the reading of the voltmeter is 10 V. Ammeter, voltmeter and battery are ideal. Find the resistance R_(3)

In the circuit given in fig. (V_C)=50 V and R=50 Omega . The values of C and (V_R) are

In the given circuit, R_(1) = 10Omega R_(2) = 6Omega and E = 10V, then correct statement is

In the circuit shown, R_1=2 Omega,R_2=R_3=10Omega, R_4=20Omega and E=6V . Work out the equivalent resistance of the circuit and the current in each resistor.

In the circuit shown, R_1=2 Omega,R_2=R_3=10Omega and E=6V . Work out the equivalent resistance of the circuit and the current in each resistor.