Basic Electrical Engineering Questions and Answers – Discharge of a Capacitor Through a Resistor

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This set of Basic Electrical Engineering Interview Questions and Answers for freshers focuses on “Discharge of a Capacitor Through a Resistor”.

1. 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
View Answer

Answer: d
Explanation: 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.
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2. 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
View Answer

Answer: c
Explanation: In a series RC circuit, the initial charging current is:
I=V/R = 200/(0.5*106s) = 400*10-6A = 400 microA.

3. A 8 microF 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
View Answer

Answer: a
Explanation: 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.
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4. 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
View Answer

Answer: b
Explanation: 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 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
View Answer

Answer: d
Explanation: 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.
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: a
Explanation: We know that Q/V=C. Hence the value of capacitance is 2F.
U=(Q*V)/2 = (10*5)/2 = 25 J.
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Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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