Important MCQ of RL Circuit With Explanation – 2021

Answer.1. 3.84 ∠20.6° V

Explanation:-

Inductive reactance X_L = 2πfL

X_L = 6.28 × 8 × 90

X_L = 4.5216 kΩ

Impedance in polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_L} \angle {\tan ^{ – 1}}\left( {\frac{{{X_L}}}{R}} \right)\\ \\ Z = \sqrt {{{(12)}^2} + {{(4.5216)}^2}} \angle {\tan ^{ – 1}}\left( {\frac{{4.5216}}{{12}}} \right) \end{array}$

Z = 12.82∠20.6

Source voltage = current × Impedence

V_s = 0.3 ∠0° × 12.82∠20.6

V_s = 3.84 ∠20.6° V

Answer.1. 0 W

Explanation:-

Inductive reactance X_L = 2πfL

X_L = 6.28 × 60 × 20

X_L = 7536Ω

Current Drawn I_L = V_L/X_L

I_L =110/7536 = 14.6 mA

Active power P = V.I.Cosφ

Since the resistance is zero the circuit is purely inductive in nature and phase angle φ = 90°

110(14.6 × 10⁻³ ) (cos90°)

P = 0 Watt

Answer.4. 1526 Ω

Explanation:-

Inductive reactance X_L = 2πfL

X_L = 6.28 × 10 × 15

X_L = 0.942 kΩ

Impedance Z

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_L} \\ \\ Z = \sqrt {{{(1.2)}^2} + {{(0.942)}^2}} \end{array}$

Z = 1.526 kΩ

Z = 1526 Ω

Answer.3. 0.1

Explanation:-

The power factor is the ratio of the actual power being consumed by an appliance to the apparent power it draws from the source.

In the case of a pure resistive circuit, the power factor is 1, and the current and voltage are in the same phase. Also in the pure resistive circuit, the component is dissipating as much power (real power) as it is drawing from the source (apparent power).

So from the above discussion, it is clear that

Power factor ∝ Energy dissipation or Energy losses.

Hence the minimum energy loss will occur when the power factor is 0.1.

Answer.3. 12.18 ∠-76° mS

Explanation:-

Conductance G = 1/R = 1/470 = 2.13 mS

Inductive reactance X_L = 2πfL

X_L = 6.28 × 1.5 × 200

X_L = 1884 Ω

Inductive susceptance B_L

B_L = 1/X_L = 1/1884 = 0.5 ms

The total admittance

$\begin{array}{l} Y = \sqrt {{G^2} + {B^2}_L} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_L}}}{G}} \right)\\ \\ {\rm{Y = }}\sqrt {{{(2.12)}^2} + {{(0.5)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{0.5}}{{2.12}}} \right) \end{array}$

Y = 12.18 ∠-76° mS

Answer.2. 4.24 A

Explanation:-

In a parallel RL circuit, the current is additive

$\begin{array}{l} {I_{rms}} = \sqrt {{I^2}R + {I^2}L} \\ \\ {I_{rms}} = \sqrt {{{(3)}^2} + {{(3)}^2}} \end{array}$

I_RMS = 4.24 A

Answer.2. Decrease

Explanation:-

The resistor and the inductor always share the input voltage. Thus, the resistor too may absorb a part of the input voltage in the circuit. Therefore, the voltage available to the inductor during this interval can only be less than what is available in the source. 

• As the resistor voltage tends to increase in the RL circuit the inductor voltages decrease therefore the current increases as shown in the graph.
• When the inductor voltage becomes zero then the voltage source output will be across the resistor, and the current will be at its maximum value.

Inductive reactance X_L = 2πfL

Hence from the above equation, the inductive reactance is directly proportional to the frequency hence to increase the current we need to decrease the frequency so that the circuit becomes more resistive in nature.

Answer.3. 26 V

Explanation:-

In series RL circuit the voltage is additive

$\begin{array}{l} {E_{rms}} = \sqrt {{V^2}_R + {V^2}_L} \\ \\ {E_{rms}} = \sqrt {{{(12)}^2} + {{(14)}^2}} \end{array}$

E_RMS = √340

E_Max = E_RMS × √2

E_Max = √680

E_Max = 26.0 V

Answer.1. 8.74 mA

Explanation:-

Inductive reactance X_L = 2πfL

X_L = 6.28 × 2000 × 120

X_L = 1.5072 kΩ

${\rm{Z = }}\frac{{R{X_L}}}{{\sqrt {{R^2} + {X^2}_L} }}$

${\rm{ = }}\frac{{(1.50)(3.3)}}{{\sqrt {{{(3.3)}^2} + {{(1.50)}^2}} }}$

Z = 1.367 kΩ

The total current in the circuit I

I = V/Z = 12/1.367

I = 8.74 mA

Answer.2. 120 mA

Explanation:-

Since the voltage source is parallel to both resistances.

So, In parallel circuit V₁ = V₂ = V₃ = V

Current in inductor, I = V/X_L =15/125 = 120 mA

Answer.4. 115∠60.8°Ω

Explanation:-

The impedance in polar form is

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

${\rm{ = }}\sqrt {{{56}^2} + {{100}^2}} \angle {\tan ^{ – 1}}\left( {\dfrac{{100}}{{56}}} \right)$

115∠60.8°Ω