Q. 70

# How many time constants will elapse before the current in a charging RC circuit drops to half of its initial value? Answer the same question for a discharging RC circuit.

Answer :

Current when capacitor is discharging:

A capacitor of capacitance C is being discharged through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I0 is the initial current.

Note that is known as time constant

Current when capacitor is charging:

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. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:

Where I_{0} is the initial current.

Note that is known as time constant.

In both the cases wish to find time t>0 such that

I(t) is given by the same formula in both the cases.

Hence, 0.69 time constants will elapse.

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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