Basic Electrical Engineering Questions and Answers
a) 7.08*10-11F
b) 7.08*1011F
c) 7.08*10-12F
d) 7.08*10-10F
Answer: a
The expression to find capacitance when a dielectric is introduced between the plates is: C=ke0A/d. Substituting the given values in the equation, we get C = 7.08*10-11F.
Q.2
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
Answer: b
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.
Q.3
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
Answer: c
We know that: Q=Q0(1-e-t/RC). When RC=t, we have: Q=Q0(1-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.
Q.4
An 8microF 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
Answer: d
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.
Q.5
An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the initial charging current.
a) 100 microA
b) 500 microA
c) 400 microA
d) 1000microA
Answer: c
In a series RC circuit, the initial charging current is:
I=V/R = 200/(0.5*106s) = 400*10-6A = 400 microA.
Q.6
An 8microF 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
Answer: b
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.
Q.7
An 8microF 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 microA
b) 68 microA
c) 48 microA
d) 74 microA
Answer: d
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 microA.
Q.8
If the switch is closed at t=0, what is the current in the circuit?
b) 10A
c) 20A
d) Infinity
a) 0V
b) 10V
c) 20V
d) Infinity
Answer: a
When the switch is closed at t=0, the capacitor has no voltage across it since it has not been charged. The capacitor acts as a short circuit and the voltage across it is zero.
Q.10
Calculate di(0)/dt if the switch is closed at t=0.
b) -10A/s
c) 0A/s
d) -0.1A/s
Answer: d
Applying KVL to the given circuit, we get:
i=i0e-t/RC
= (100/10)e-t/100
i=10 e-t/100
di/dt = -(10/100) e-t/100
di(0)/dt=-0.1A/s.
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