If the power factor is `1//2` in a series `RL` circuit with `R = 100 Omega`. If `AC` mains, `50 Hz` is used then `L` is

If power factor is 1//2 in a series RL, circuit R=100 Omega. AC mains is used then L is

In a series L-C-R circuit R = 100 Omega , L = 1H and C= 1 mu E The half power bandwidth is ……..

Assertion (A) If the frequency of the applied AC is doubled , then the power factor of a series R-L circuit decreases. Reason (R ) Power factor of series R-L circuit is given by costheta=(2R)/(sqrt(R^(2)+omega^(2)L^(2)))

Assertion (A) If the frequency of the applied AC is doubled , then the power factor of a series R-L circuit decreases. Reason (R ) Power factor of series R-L circuit is given by costheta=(2R)/(sqrt(R^(2)+omega^(2)L^(2)))

A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R=6 Omega , L=50.96 mH and C=398 muF . Find the power dissipated in the circuit

An a.c. source is connected across an LCR series circuit with L = 100 mH, C = 0.1muF and R = 50 Omega . The frequency of ac to make the power factor of the circuit, unity is

An a.c. source is connected across an LCR series circuit with L = 100 mH, C = 0.1muF and R = 50 Omega . The frequency of ac to make the power factor of the circuit, unity is

The power factor of an L-R series circuit is 0.5 and that of a C-R series circuit is 0.2. If the elements (L, C and R) of the two circuit are joined in series and connected to the same ac source, the power factor of this circuit is found to be 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is

In a series RC circuit with an AC source, R = 300 Omega, C = 25 muF, epsilon_0= 50 V and v = 50/(pi) Hz . Find the peak current and the average power dissipated in the circuit.

In a series RC circuit with an AC source, R = 300 Omega, C = 25 muF, epsilon_0 50 V and v = 50/((Pi) Hz . Find the peak current and the average power dissipated in the circuit.