Basic Electrical Engineering Multiple Choice Questions 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
Answer: D
Clarification: XL=2*π*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
Answer: A
Clarification: 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
Answer: B
Clarification: XL=2*π*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
Answer: B
Clarification: XL=2*π*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
Answer: C
Clarification: XL=2*π*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)
Answer: A
Clarification: 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
Answer: B
Clarification: 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
Answer: C
Clarification: 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
Answer: B
Clarification: In Impedance triangle, Base is R, Hypotenuse is Z, Height is XL.
So, sinϕ = XL/Z.