Home
Class 12
PHYSICS
In the LCR circuit shown in figure unkno...

In the LCR circuit shown in figure unknown resistance and alternating voltage source are connect Then switch 'S' is closed then there is a phase difference of `(pi)/(4)` between current and applied voltage and voltage accross resister `(100)/(sqrt2)V` is phase. Neglecting resistance of connecting wire answer the following questions:

Average power consumption in the circuit when 'S' is open

A

2500W

B

3000W

C

5000W

D

1250W

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    RESONANCE|Exercise EXERCISE|10 Videos
  • ALTERNATING CURRENT

    RESONANCE|Exercise PART-II|6 Videos
  • ALTERNATING CURRENT

    RESONANCE|Exercise PART III:|15 Videos
  • ATOMIC PHYSICS

    RESONANCE|Exercise Advanved level problems|17 Videos

Similar Questions

Explore conceptually related problems

In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected.When swith S is closed then there is a phase difference of pi/4 between current and applied voltage and voltage across resister is 100/sqrt2 V .When switch is open current and applied voltage are in same phase.Neglecting resistance of connecting wire answer the following questions : Peak voltage of applied voltage sources is:

In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected.When swith S is closed then there is a phase difference of pi/4 between current and applied voltage and voltage across resister is 100/sqrt2 V .When switch is open current and applied voltage are in same phase.Neglecting resistance of connecting wire answer the following questions : Resonance frequency of 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

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

In a series C-R circuit shown in figureure, the applied voltage is 10 V and the voltage across capacitor is found to 8 V . The voltage across R , and the phase difference between current and the applied voltage will respectively be .

In a LCR series circuit, if the phase difference between applied voltage and current is zero then

In a series LCR circuit resistance R = 10Omega and the impedance Z = 10 Omega The phase difference between the current and the voltage is

An Ac voltage V=V_0 sin 100t is applied to the circuit, the phase difference between current and voltage is found to be pi/4 , then .