# Discharge of Capacitor Through Resistor MCQ [Free PDF]

1. An 8µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time constant.

A. 1s
B. 2s
C. 3s
D. 4s

The time constant is the product of the resistance and capacitance in a series RC circuit.

Therefore, time constant = 8 × 10-6 × 4 × 106 = 4s.

2. An 8µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the initial charging current.

A. 100 µn
B. 500 µA
C. 400 µA
D. 1000µA

In a series RC circuit, the initial charging current is:

I = V/R = 200/(0.5 × 106s)

= 400 × 10-6A = 400 µA.

3. A 8 µ F capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time taken for the potential difference across the capacitor to grow to 160V.

A. 6.93s
B. 7.77s
C. 2.33s
D. 3.22s

From the previous explanations, we know that the initial current is 400mA, and the time constant is 4s.

Substituting the values of capacitor voltage, initial voltage, initial current and time constant in the equation:

V = V0(1-e-t/RC)

Substituting V = 160V,

V0 = 200V,

RC = 4s we get,

t = 6.93s.

4. An 8µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the voltage in the capacitor 4s after the power is supplied.

A. 123.4V
B. 126.4V
C. 124.5V
D. 132.5V

We can get the value of the potential difference across the capacitor in 4s, from the following equation:

Vc = V(1-e-t /RC).

Substituting the values in the given equation, we get Vc = 126.4V.

5. An 8µF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the current in the capacitor 4s after the power is supplied.

A. 79 µA
B. 68 µA
C. 48 µA
D. 74 µA

In the given question, the time constant is equal to the time taken = 4s.

Hence the value of current will be 37% of its initial value

= I = 0.37 × 200 = 74 µA.

6. A circuit has a resistance of 2 ohms connected in series with a capacitance of 6F. Calculate the discharging time constant.

A. 3
B. 1
C. 12
D. 8

The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance

= 2 × 6 = 12.

7. What is the energy in a capacitor if the voltage is 5V and the charge is10C?

A. 25J
B. 35J
C. 54J
D. 55J

We know that Q/V = C.

Hence the value of capacitance is 2F.

U = (Q × V)/2 = (10 × 5)/2 = 25 J.

8. A CR network is one which consists of _________

A. A capacitor and resistor connected in parallel
B. A capacitor and resistor connected in series
C. A network consisting of a capacitor only
D. A network consisting of a resistor only

A CR network consists of a capacitor connected in series with a resistor. The capacitor discharges or charges through the resistor.

9. At DC, capacitor acts as _________

A. Open circuit
B. Short circuit
C. Resistor
D. Inductor

Capacitive Reactance XC  = 1/(2πfC).

For DC, f = 0 so, XC becomes infinite.

Hence for dc, the capacitor acts as an open circuit.

10. In an RC series circuit, when the switch is closed and the circuit is complete, what is the response?

A. Response does not vary with time
B. Decays with time
C. Increases with time
D. First increases, then decrease