D. C. Pandey Solutions for Chapter: Alternating Current, Exercise 13: [ Level 2 ]

Author:D. C. Pandey

D. C. Pandey Physics Solutions for Exercise - D. C. Pandey Solutions for Chapter: Alternating Current, Exercise 13: [ Level 2 ]

Attempt the free practice questions on Chapter 6: Alternating Current, Exercise 13: [ Level 2 ] with hints and solutions to strengthen your understanding. Complete Study Pack for Engineering Entrances Objective Physics Vol 2 solutions are prepared by Experienced Embibe Experts.

Questions from D. C. Pandey Solutions for Chapter: Alternating Current, Exercise 13: [ Level 2 ] with Hints & Solutions

HARD
JEE Advanced
IMPORTANT

An AC voltage V=V0sin100 t is applied to the circuit, the phase difference between current and voltage is found to be π4, then

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MEDIUM
JEE Advanced
IMPORTANT

When an alternating voltage of 220 V is applied across a device P, a current of 0.25 A Flows through the circuit, and it leads the applied voltage by an angle π2 radian . When the same voltage source is connected across another device Q, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of P and Q.

HARD
JEE Advanced
IMPORTANT

An inductor XL=2 Ω a capacitor XC=8 Ω and a resistance R=8 Ω are connected in series with an AC source. The voltage output of AC source is given by V=10cos100πt

Question Image

The instantaneous potential difference between points A and B, when the applied voltage is 3th5 of the maximum value of applied voltage is

HARD
JEE Advanced
IMPORTANT

A group of electric lamps having a total power rating of 1000 W is supplied by an AC voltage E=200sin310t+60° Then, the rms value of the circuit current is 

HARD
JEE Advanced
IMPORTANT

One 10 V, 60 W bulb is to be connected to a 100 V line. The required self-inductance of the inductor coil will be: f=50 Hz

MEDIUM
JEE Advanced
IMPORTANT

An alternating voltage V=30sin 50t+40cos 50t is applied to a resistor of resistance 10 Ω. The rms value of the current through resistor is

MEDIUM
JEE Advanced
IMPORTANT

Assertion: In L-C-R series AC circuit  XL=XC=R at a given frequency. When this frequency is doubled, the impedance of the circuit is 132R.

Reason: The given frequency is resonance frequency.

EASY
JEE Advanced
IMPORTANT

Assertion: At resonance power factor of series L-C-R circuit is zero.

Reason: At resonance current function and voltage functions are in the same phase.