Question

A fully charged capacitor of 10µF has a potential difference of 100V across its terminals. It is discharged through 1 K Ohm resistor. Find time constant of the circuit.

Answer 1

Explanation:

Answer

Open in answr app

Correct option is

A

its time constant is 1×10−1s

B

during time = 0.693 ×10−1s, the charge left on the capacitor is half of its maximum charge

D

during time = its time constant (τ), charge left on the capacitor is 0.37 times the maximum charge

We know that the formula of charge on a capacitor is -

q=qo(e−t/τ) while discharging

Where time constant τ=RC=100×106×1000=0.1s=10−1s

and qo is maximum charge.

When charge on capacitor is half of maximum charge-

2qo=qo(e−t/τ)

⟹e−t/τ=21

⟹t=τln2=0.693×10−1s

For time t=τ, charge on capacitor is-

q=qo(e−t/τ)=qoe−1

⟹q=0.37qo

That is charge on capacitor is 0.37 times maximum charge.

Answer-(A),(B),(D)

Answer 2

Answer:

The time constant of the circuit is 10 millisecond.

Explanation:

Time constant

• The time constant of a circuit is the time required to discharge the capacitor to 1/e of value of initial voltage.
• The time required to discharge the capacitor's voltage to V/e from V is time constant. Value of e is 2.718. So, 36.79% of V is the voltage at time constant.

If capacitor is attached to the resistance then capacitor starts to discharge. In this RC circuit the value of time constant τ is given by

τ = RC

where R is the value of resistance and C is the capacitance of the capacitor.

The given value of capacitance, C = 10 μF.

The given value of resistance, R = 1000 Ω.

So, time constant τ is given by

τ = RC (1)

Putting values of R and C in equation (1),

τ = 10 μF × 1000 Ω

τ = 10 ms.

So, the value of time constant is given by 10 millisecond.