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*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 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*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 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) 57.4V
b) 42.6V
c) 65.2V
d) 76.2V
View Answer
Explanation: XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.
Voltage across resistor= 8.2*7=57.4V.
Voltage across inductor =100-VR= 42.6V.
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*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R2+XL2)=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: Form 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: Form the voltage triangle, we get cosφ= VR/V.
Hence φ=cos-1VR/V.
9. An RL network is one which consists of?
a) Resistor and capacitor in parallel
b) Resistor and capacitor in series
c) Resistor and inductor in parallel
d) Resistor and inductor in series
View Answer
Explanation: An R-L network is a network which consists of a resistor which is connected in series to an inductor.
10. At DC, inductor acts as ___________
a) Open circuit
b) Short circuit
c) Resistor
d) Inductor
View Answer
Explanation: At DC, the inductor acts as short circuit because the inductive resistance is zero. The frequency of a DC circuit is 0. The inductive resistance=(2*pi*f*L). Therefore, if the frequency is 0, the inductive resistance is zero and it acts as an short circuit.
Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.
To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
