This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Growth and Decay”.

1. The charging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________% of the initial charge.

a) 33

b) 63

c) 37

d) 36

View Answer

Explanation: We know that: Q=Q0(1-e

^{-t}/RC).

When RC=t, we have: Q=Q0(1-e

^{-1})= 0.63*Q0.

Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.63 times its initial charge.

2. The discharging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________% of the initial charge.

a) 33

b) 63

c) 37

d) 36

View Answer

Explanation: We know that: Q=Q0(e

^{-t}/RC).

When RC=t, we have: Q=Q0(e

^{-1})= 0.37*Q0.

Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.37 times its initial charge.

3. A circuit has a resistance of 2 ohm connected in series with a capacitance of 6F. Calculate the charging time constant.

a) 3

b) 1

c) 12

d) 8

View Answer

Explanation: The charging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance= 2*6=12.

4. A circuit has a resistance of 5 ohm connected in series with a capacitance of 10F. Calculate the discharging time constant.

a) 15

b) 50

c) 5

d) 10

View Answer

Explanation: The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance= 5*10=50.

5. What is the value of current in a discharging capacitive circuit if the initial current is 2A at time t=RC.

a) 0.74A

b) 1.26A

c) 3.67A

d) 2.89A

View Answer

Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 37% of the initial charge in the discharging circuit. Hence, I=2*0.37= 0.74A.

6. What is the value of current in a charging capacitive circuit if the initial current is 2A at time t=RC.

a) 0.74A

b) 1.26A

c) 3.67A

d) 2.89A

View Answer

Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 63% of the initial charge in the charging circuit. Hence, I=2*0.63= 1.26A.

7. While discharging, what happens to the current in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

View Answer

Explanation: The equation for the value of current in a discharging capacitive circuit is:

I=I0*e

^{-t}/RC. From this equation, we can see that the current is exponentially decreasing since e is raised to a negative power.

8. While discharging, what happens to the voltage in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

View Answer

Explanation: The equation for the value of voltage in a discharging capacitive circuit is:

V=V0*e

^{-t}/RC. From this equation, we can see that the voltage is exponentially decreasing since e is raised to a negative power.

9. While charging, what happens to the current in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

View Answer

Explanation: The equation for the value of the current in a charging capacitive circuit is:

I=I0*(1-e

^{-t}/RC). From this equation, we can see that the current is exponentially increasing since e is raised to a negative power and we are subtracting it from 1. Hence as the value of e

^{-t}/RC increases, the current increases exponentially.

10. While charging, what happens to the voltage in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

View Answer

Explanation: The equation for the value of voltage in a charging capacitive circuit is:

V=V0*(1-e

^{-t}/RC). From this equation, we can see that the voltage is exponentially increasing since e is raised to a negative power and we are subtracting it from 1. Hence as the value of e

^{-t}/RC increases, the voltage increases exponentially.

**Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.**

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