This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Resistance and Inductance in Series”.
1. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.
a) 2.2A
b) 4.2A
c) 6.2A
d) 8.2A
View Answer
Explanation: XL=2*Ï€*f*L = 10 ohm. Z2=(R2+XL2)
Therefore the total impedance Z = 12.2ohm.
V=IZ, therefore I=V/Z=100/12.2 = 8.2A.
2. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the phase difference.
a) -55.1
b) 55.1
c) 66.1
d) -66.1
View Answer
Explanation: φ=tan-1(XL/R)=55.1
Since this is an inductive circuit, the current will lag, hence φ= -55.1.
3. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the resistor.
a) 31.8V
b) 57.4V
c) 67.3V
d) 78.2V
View Answer
Explanation: XL=2*Ï€*f*L = 10 ohm. Z2=(R2+XL2)
Therefore, the total impedance Z = 12.2ohm.
V=IZ, therefore I=V/Z=100/12.2 = 8.2A. Voltage across resistor = 8.2*7 = 57.4V.
4. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the inductor.
a) 52V
b) 82V
c) 65V
d) 76V
View Answer
Explanation: XL=2*Ï€*f*L = 10 ohm. Z2=(R2+XL2)
Therefore, the total impedance Z = 12.2ohm.
V=IZ, therefore I=V/Z=100/12.2 = 8.2A. Voltage across inductor = 8.2*10 = 82V.
5. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a x V 50Hz sinusoidal supply. The current in the circuit is 8.2A. Calculate the value of x.
a) 10V
b) 50V
c) 100V
d) 120V
View Answer
Explanation: XL=2*Ï€*f*L= 10 ohm. Z2=(R2+XL2)
Therefore, the total impedance Z = 12.2ohm.
V=IZ, therefore V = 12.2*8.2 = 100V.
6. Which, among the following, is the correct expression for φ.
a) φ=tan-1 (XL/R)
b) φ=tan-1 (R/XL)
c) φ=tan-1 (XL*R)
d) φ=cos-1 (XL/R)
View Answer
Explanation: From the impedance triangle, we get tanφ= XL/R.
Hence φ=tan-1 (XL/R).
7. For an RL circuit, the phase angle is always ________
a) Positive
b) Negative
c) 0
d) 90
View Answer
Explanation: For a series resistance and inductance circuit the phase angle is always a negative value because the current will always lag the voltage.
8. What is φ in terms of voltage?
a) φ=cos-1V/VR
b) φ=cos-1V*VR
c) φ=cos-1VR/V
d) φ=tan-1V/VR
View Answer
Explanation: From the voltage triangle, we get cosφ= VR/V.
Hence φ=cos-1VR/V.
9. What is sinϕ from impedance triangle?
a) XL/R
b) XL/Z
c) R/Z
d) Z/R
View Answer
Explanation: In Impedance triangle, Base is R, Hypotenuse is Z, Height is XL.
So, sinϕ = XL/Z.
Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.
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