Home
Class 12
PHYSICS
A resistor of 500 Omega and an inductanc...

A resistor of `500 Omega` and an inductance of 0.5 H are in series with an ac source which is given by `V=100sqrt(2)sin(1000t)`. The power factor of the combination is

A

`(1)/(sqrt(2))`

B

`(1)/(sqrt(3))`

C

`0.5`

D

`0.6`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    NCERT FINGERTIPS|Exercise Lc Oscillations|9 Videos

  • ALTERNATING CURRENT

    NCERT FINGERTIPS|Exercise Transformers|16 Videos

  • ALTERNATING CURRENT

    NCERT FINGERTIPS|Exercise Ac Voltage Applied To A Series Lcr Circuit|28 Videos

  • ATOMS

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

A capacitor of 100 muF , a resistor of 20 Omega and an inductor of inductance L are connected in series with an a.c. source of frequency 50 Hz. Calculate the value of inductance L of the inductor, if phase angle between current and voltage is zero.

A coil having a resistance of 50.0 Omega and an inductance of 0.5 H is connected to an AC source of 110 , v, 50Hz . The impedance of the coil is (appoximately)

A 200 Omega resistor and 1H inductor are joined in series with an ac source of emf 10 sqrt(2) sin (200t)V . Calculate the phases difference between emf and current.

An inductor 200 mH, capacitor 500mu F and resistor 10 Omega are connected in series with a 100 V variable frequency ac source. What is the frequency at which the power factor of the circuit is unity?

A coil having a resistance of 50.0 Omega and an inductance of 0.5 H is connected to an AC source of 110 , v, 50Hz . If the frequency of the AC is decreased by a factor of 2 , then the rms current

A coil having a resistance of 50.0 Omega and an inductance of 0.5 H is connected to an AC source of 110 , v, 50Hz . The rms current in the circuit is

In an L-R circuit, an inductance of 0.1 H and a resistance of 1 Omega are connected in series with an ac source of voltage V = 5 sin 10 t. The phase difference between the current and applied voltage will be