List of top Physics Questions on Current Electricity

In an LCR series circuit, the source of emf is E=30sin(100t), R=120Ω, L=100mH, C=100μF.
(A) The numerical value of impedance 
(B) The numerical value of resistance R 
(C) The numerical value of capacitive reactance 
(D) The numerical value of inductive reactance.
Arrange the values of quantities mentioned in (A, B, C, D) in increasing order.

A long solenoid is formed by winding insulated copper wire at the rate of 20 turns per cm. The current required to produce a magnetic field of 20 mT inside the solenoid at its center would be:

For a series LCR circuit, at the condition of resonance, the value of the power factor will be:

A capacitor and a coil with resistance R are in series and connected to a 6 V AC source. By varying the frequency of the source, a maximum current of 600 mA is observed. If the same coil is now connected to a cell of emf 6 V and internal resistance of 2 ohms, the current through it will be:

Arrange the following materials in the increasing order of their resistivity: (A) Copper, (B) Platinum, (C) Silver, (D) Aluminium.

In the circuit shown, the current in the 1Ω resistor is:

The current sensitivity of a moving coil galvanometer is doubled by making the number of turns double. Then its voltage sensitivity will be:

In a coil, an increase in current from 5 A to 10 A in 100 ms induces an emf of 100 V. The self-inductance of the coil is:

A conducting wire of length L, uniform area of cross-section A, and material having n free electrons per unit volume offers a resistance R to the flow of current. m and e are the mass and charge of an electron, respectively. If τ is the mean free time of the electrons in the conductor, the correct formula for resistance R is:

A galvanometer with resistance 100 Ω gives full-scale deflection with a current of 2 mA. The resistance required to convert the galvanometer into an ammeter of range 0 to 20 A is nearly:

An electric bulb rated 200 V - 100 W is connected to a 160 V power supply. The power consumed by it would be:

In the given circuit find the voltage across 12 resistance.

A galvanometer of 50 \(\Omega\) resistance has 25 division. A current of \(4×10^{-4} A\) gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of: