Answer.4. Can’t be determined

Explanation:-

The impedance of the RL circuit is given as

$Z = \sqrt {{R^2} + {\omega ^2}{L^2}} $

Now the resistance is doubled and the frequency is halved

R = 2R’

&

ω’= ω/2

∴ $Z’ = \sqrt {4{R^2} + \frac{{{\omega ^2}{L^2}}}{4}} $

Hence without knowing the exact value of resistance R, Frequency F, and Inductance L we can’t determine the Impedance Z.

Answer.2. Increase

Explanation:-

The phase angle of an RL circuit is given as

φ = tan⁻¹(ωL/R)

φ = tan⁻¹(2πfL/R)

From the above equation the frequency f ∝ Phase angle φ

By increasing the frequency in the RL circuit the phase angle φ also increases.

Answer.1. Decrease

Explanation:-

Variation of Impedance and Phase Angle with Frequency

The impedance triangle is important to understand that how the frequency of the applied voltage affects the RL circuit response. The inductive reactance varies directly with frequency. When X_L increases, the magnitude of the total impedance also increases; and when X_L decreases, the magnitude of the total impedance decreases. Thus, Z is directly dependent on frequency.

The phase angle φ also varies directly with frequency because φ = tan⁻¹(X_L/R). As X_L increases with frequency, so does φ, and vice versa.

The impedance triangle illustrates the variations in X_L, Z, and θ as the frequency changes. Of course, R remains constant. The main point is that because X_L varies directly with the frequency, so also do the magnitude of the total impedance and the phase angle.

Answer.4. Resistance and Inductive Reactance

Explanation:-

The impedance of the RL circuit is given as

$Z = \sqrt {{R^2} + {\omega ^2}{L^2}} $

From the above equation, it is clear that the Impedance Z is the summation of both Resistance (R) and inductive Reactance XL.

$X = \sqrt {{R^2} + {X^2}_L} $

Answer.2. Increase

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.
• The phase angle is defined as the angle between effective resistance and effective reactance of a circuit.

Cosφ = R/Z.

V = IZ

∴ V of inductor directly proportional to the reactance of inductor.

• When Voltage across the inductor is greater than the voltage across Resistor. Then Inductive reactance is greater than the resistance then the circuit becomes inductive in nature and the phase angle increases.

φ = tan⁻¹(X_L/R)

• In short, if the resistor voltage in a series RL circuit becomes less than the inductor voltage, then the phase angle increases.

Answer.2. 0.564

Explanation:-

Given Resistance R = 1.5 kΩ

Inductive reactance X_L = 2.2 kΩ

The phase angle of an RL circuit is given as

φ = tan⁻¹(X_L/R)

φ = tan⁻¹(2.2/1.5)

φ = 55.59°

Power factor = COSφ = C0s(55.59°)

Power factor = 0.564

Answer.4. 14.6 mA

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

Answer.1. 55.15 Ω

Explanation:-

The Impedance of the Parallel RL circuit is given as

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

$\frac{{(140)(60)}}{{\sqrt {{{(140)}^2} + {{(60)}^2}} }}$

Z= 55.15Ω

Answer.2. 12 VA

Explanation:-

Active power is given as

P = V.I.Cosφ

In the case of a pure inductive circuit, the power angle is φ = π/2 or 90°.

P = V.I.Cos90°

Thus the active power is zero in the case of a pure inductive circuit and the reactive power is maximum.

The combination of reactive power and true power or active power is called apparent power.

Apparent power = Active power + Reactive power

In this case, active power is zero.

Apparent power = 0 + 12

Apparent power = 12 VA

Answer.1. 129 ∠68.6° Ω

Explanation:-

The impedance of RL series circuit in polar form is given as

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_L} \angle {\tan ^{ – 1}}\left( {\frac{{{X_L}}}{R}} \right)\\ \\ Z = \sqrt {{{(47)}^2} + {{(120)}^2}} \angle {\tan ^{ – 1}}\left( {\frac{{120}}{{47}}} \right) \end{array}$

Z = 129 ∠68.6° Ω