Home
Class 12
PHYSICS
In series L-R circuit, XL=R. Now a capac...

In series `L-R` circuit, `X_L=R`. Now a capacitor with `X_C=R` is added in series. New power factor

A

Same as initial

B

`1/sqrt2` times the initial

C

`1/2` times the initial

D

`sqrt2` times the initial

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

In series LR circuit, X_(L) = 3 R . Now a capacitor with X_(C ) = R is added in series. The ratio of new to old power factor

In L-C-R series circuit

In L-C-R series AC circuit,

In a series LR circuit X_(L) = R and power factor of the circuit is P_(1) . When capacitor which capacitance C such that X_(L) =X_(C) is put in series , the power factor becomes P_(2) .Calculate P_(1)//P_(2)

In a series L.R circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capactor of value C is added in series to the L and R. Taking the value of C as (n/(3pi))muF , then value of n is ____________

If V_0 and I_0 are the peak current and voltge across the resistor in a series L-C-R circuit, then the power dissipated in the circuit is (power factor =costheta)

In the L-C-R circuit as shown in figure,

A series L-C-R circuit is operated at resonance . Then

In a series R, L, C circuit X_(L)=10Omega, X_(c) = 4omega and R = 6Omega . Find the power factor of the circuit

In an L-C-R series, AC circuit at resonance