Alternating Cur...

In a series resonant L-C-R circuit, the voltage across R is 100 V and R =1 kΩ with C = 2µF. The resonant frequency co is 200 rad s^-1. At resonance, the voltage across L is

A fully charged capacitor C with initial charge q₀ is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is

At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in sec) does the voltage across them become 4 V? (take, ln 5 =1.6, ln 3 =1.1)

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

An inductor of 2 H and a resistance of 10Ω are connected in series with a battery of 5 V. The initial rate of change of current is

The time constant of the given circuit is

An inductor (L=100 mH) a resistor (R =100Ω) and a battery (E =100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is

Find the time in which current becomes 81 mA, when the switch S is closed at t= 0.

Current from A to B in the straight wire is decreasing. The direction of induced current in the loop, Fig. is