Ques.22. In RL lead circuit the phase angle φ is given as
φ = tan−1(XL/R)
φ = tan(XL/R)
φ = tan−1(R/XL)
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 )
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 = IXL (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
60.2°
65.4°
75.2°
71.6°
Answer.4. 71.6°
Explanation:-
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
φ = tan−1(XL/R)
φ = tan(XL/R)
φ = tan−1(R/XL)
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.
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
16.4°
20.6°
18.4°
12.24°
Answer.3. 18.4°
Explanation:-
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.
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).
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.
4.12 − j1.64
3.03 − j1.59
3.12 − j1.64
3.25 − j1.59
Answer.2. 3.03 − j1.59
Explanation:-
First, determine the conductance magnitude R = 330Ω. thus,