Q. 645.0( 1 Vote )

# A 20 μF capacitor is joined to a battery of emf 6.0 V through a resistance of 100 Ω. Find the charge on the capacitor 2.0 ms after the connections are made.

Answer :

Concepts/Formulas used:

Charging a capacitor:

A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed

The charge is at any t>0 is given by:

Note that is known as time constant.

Given,

Capacitance,

EMF of battery,

Resistance,

Now,

Also,

We know that,

At t = 2.0ms,

Hence, the charge on the capacitor at t = 2.0ms is 76μC.

Rate this question :

A parallel-plate capacitor has plate area 20 cm^{2}, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?

HC Verma - Concepts of Physics Part 2