Important MCQ of RL Circuit With Explanation – 2021

Ques.21. The current in Figure is expressed in polar form as mA. What will be the source voltage in the RL series circuit expressed in polar form?

  1. 4.84∠32.1° V
  2. 2.36∠32.1° V
  3. 236∠32.1° V
  4. 23.6∠32.1° V

Answer.2. 2.36∠32.1° V

Explanation:-

RL circuit.2

From the given figure

Resistance = 100 kΩ

Inductance = 100 mH

Frequency = 10 kHz

The magnitude of the inductive reactance is

XL = 2πfL

XL = 2π × 10 × 100 = 6.28 kΩ

The impedance in rectangular form is

Z = R + jXL = 10kΩ + j6.28kΩ

Converting to polar form

$\begin{array}{l} {\rm{Z = }}\sqrt {{R^2} + {X^2}_L} \angle {\tan ^{ – 1}}\left( {\dfrac{{{X_L}}}{R}} \right)\\ \\ {\rm{ = }}\sqrt {{{(10)}^2} + {{(6.28)}^2}} \angle {\tan ^{ – 1}}\left( {\dfrac{{6.28}}{{10}}} \right) \end{array}$

 

Z = 11.8∠32.1°kΩ

Use Ohm’s law to determine the source voltage

Vs = IZ

= (0.20°mA) × (11.8∠32.1°kΩ)

Vs = 2.36∠32.1° V

 

Ques.22. In RL lead circuit the phase angle φ is given as

  1. φ = tan−1(XL/R)
  2. φ = tan(XL/R)
  3. φ = tan−1(R/XL)
  4. φ = tan(R/XL)

Answer.3. φ = tan−1(R/XL)

Explanation:-

An RL lead circuit is a phase shift circuit in which the output voltage leads the input voltage by a specified amount. Figure 12(a) shows a series RL circuit with the output voltage taken across the inductor. In the RL lead circuit (Vout = VL )

RL Lead circuit

Since VL leads by 90°, the phase angle between the inductor voltage and the input voltage is the difference between 90°.

Phase Difference Between Input and Output

The angle between Vout and Vin is given by φ.

The polar expressions for the input voltage and the current are Vin∠0° and I∠−θ respectively. The output voltage in polar form is

Vout = (I∠−θ) × (XL∠90°)

Vout = IX(90° − θ)

This expression shows that the output voltage is at an angle of (90° − θ)  with respect to the input voltage. Since θ =tan−1(XL/R), the angle between the input and output is

φ = 90° − tan−1(XL/R)

Equivalently, this angle can be expressed as

φ = tan−1(R/XL)

 

Ques.23. Determine the amount of phase lead from input to output in each lead circuit as shown in RL  circuit

  1. 60.2°
  2. 65.4°
  3. 75.2°
  4. 71.6°

Answer.4. 71.6°

Explanation:-

RL circuit.3

From the given figure

Resistance = 15 kΩ

Inductance = 5 kΩ

For the RL  lead circuit, the phase difference between Input and Output will be

φ = tan−1(R/XL)

= tan−1(15/5) = 71.6°

 

Ques.24. In RL lead circuit the phase angle φ is given as

  1. φ = tan−1(XL/R)
  2. φ = tan(XL/R)
  3. φ = tan−1(R/XL)
  4. φ = tan(R/XL)

Answer.1. φ = tan−1(XL/R)

Explanation:-

An RL lag circuit is a phase shift circuit in which the output voltage lags the input voltage by a specified amount. When the output of a series RL circuit is taken across the resistor rather than the inductor, as shown in Figure, it becomes a lag circuit.

RL Lag circuit

Phase Difference Between Input and Output in RL Lag circuit

In a series RL circuit, the current lags the input voltage. Since the output voltage is taken across the resistor, the output lags the input, as indicated by the phasor diagram in Figure.

As in the lead circuit, the amount of phase difference between the input and output and the magnitude of the output voltage in the lag circuit is dependent on the relative values of the resistance and the inductive reactance. When the input voltage is assigned a reference angle of 0°, the angle of the output voltage φ with respect to the input voltage equals θ because the resistor voltage (output) and the currents are in phase with each other. The expression for the angle between the input voltage and the output voltage is

φ = −tan−1(XL/R)

This angle is negative because the output lags the input.

 

Ques.25. Calculate the output phase angle for the RL phase Lag circuit shown in figure

  1. 16.4°
  2. 20.6°
  3. 18.4°
  4. 12.24°

Answer.3. 18.4°

Explanation:-

RL circuit.4

From the given figure

Resistance = 15 kΩ

Inductance = 5 kΩ

For the RL  lag circuit, the phase difference between Input and Output will be

φ = −tan−1(XL/R)

= −tan−1(5/15) = −18.4°

The output lags the input by 18.4°.

 

Ques.26. For the RL Parallel circuit shown in the figure, determine the magnitude and phase angle of the total impedance.

RL circuit.5

  1. 34.7∠53.4°Ω
  2. 51.2∠21.2°Ω
  3. 44.7∠63.4°Ω
  4. 36.7∠21.2°

Answer.3. 44.7∠63.4°Ω

Explanation:-

From the given figure

Resistance = 100Ω

Inductance = 50Ω

$\begin{array}{l} {\rm{Z = }}\dfrac{{R{X_L}}}{{\sqrt {{R^2} + {X^2}_L} }}\angle {\tan ^{ – 1}}\left( {\dfrac{R}{{{X_L}}}} \right)\\ \\ {\rm{ = }}\dfrac{{(100)(50)}}{{\sqrt {{{(100)}^2} + {{(50)}^2}} }}\angle {\tan ^{ – 1}}\left( {\dfrac{{100}}{{50}}} \right) \end{array}$

Z = 44.7∠63.4°Ω

 

Ques.27. For parallel RL circuits, the phasor expression for inductive susceptance (BL) is

  1. jBL
  2. −jBL
  3. 2jBL
  4. None of the above

Answer.2. −jBL

Explanation:-

For parallel RL circuits, the phasor expression for inductive susceptance (BL) is

BL = 1/XL∠90°

BL∠−90° = −jBL

 

Ques.28. For parallel RL circuits, the phasor expression for Admittance (Y) is

  1. Y∠θ
  2. Y∠−θ
  3. Y∠2θ
  4. Y∠±θ

Answer.4. Y∠±θ

Explanation:-

For parallel RL circuits, the phasor expression for Admittance (Y) is

$Y = \dfrac{1}{{Z\angle \pm \theta }} = Y\angle \pm \theta $

 

Ques.29. For parallel RL circuits, the phasor expression for total Admittance (Y) of the circuit is

(1) $Y = \sqrt {{G^2} + {B^2}_L} $

 

(2) $Y = \sqrt {{G^2} – {B^2}_L} $

 

(3) $Y = \sqrt {{G^2} \pm {B^2}_L} $

 

(4) None of the above

Answer.1. $Y = \sqrt {{G^2} + {B^2}_L} $

Explanation:-

As with the RL circuit, the unit for conductance (G), inductive susceptance (BL ), and admittance (Y) are the siemens (S).

For the basic parallel RL circuit shown in Figure (a), the total admittance is the phasor sum of the conductance and the inductive susceptance, as shown in part (b).

Parallel RL circuit 1

Y = G − jBL

$Y = \sqrt {{G^2} + {B^2}_L} $

 

Ques.30. Determine the total admittance of the RL parallel circuit as shown in the given figure.

RL circuit.6

  1. 4.12 − j1.64
  2. 3.03 − j1.59
  3. 3.12 − j1.64
  4. 3.25 − j1.59

Answer.2. 3.03 − j1.59

Explanation:-

First, determine the conductance magnitude R = 330Ω. thus,

G = 1/R = 1/330 = 3.03 mS

Now Inductive Reactance

XL = 2πfL

XL = 2π × 1000 × 100 = 628 kΩ

The inductive susceptance magnitude is

BL = 1/XL

BL =  1/628 = 1.59 mS

The total admittance is

Yt = G − JBL

Yt = 3.03 − j1.59

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