RC Circuit Question and Answers – 2022

Ques.41. The magnitude of the output voltage of RC phase lag circuit is

(1) ${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)$

 

(2) ${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + 2{X^2}_C} }}} \right){V_{in}}$

 

(3) ${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

 

(4) ${V_{out}} = \left( {\frac{{2{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

 

Answer.3. ${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

Explanation:-

The magnitude of the Output Voltage:-  To evaluate the output voltage in terms of its magnitude, visualize the RC lag circuit as a voltage divider.

RC lag circuit

A portion of the total input voltage is dropped across the resistor and a portion across the capacitor. Because the output voltage is the voltage across the capacitor, it can be calculated using either Ohm’s law (Vout = IXC) or the voltage divider formula.

${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

The phasor expression for the output voltage of an RC lag circuit is

Vout = Vout ∠φ

 

Ques.42. What will be the output voltage in the RC lag circuit when the resistance is 680 kΩ,  capacitance is 0.1µF,  frequency is 1 kHz and the input voltage has an RMS value of 10 V?

  1. 9.20∠−23.2° V
  2. 920∠-23.2° V
  3. 9.20∠23.2° V
  4. 920∠23.2° V

Answer.1. 9.20∠−23.2° V

Explanation:-

Capacitive reactance

XC = 1/2πfC

XC = 1/6.28 × 1 × 0.1

XC = 1.59

The phase of an RC lag circuit is

φ = −tan−1(R/XC)

= −tan−1(680/1.59)

φ = −23.2°

The output voltage in phasor form is

Vout = Vout ∠φ

$\begin{array}{l} {V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}\angle \phi \\ \\ \\ = \left( {\frac{{1.59}}{{\sqrt {{{(680)}^2} + {{(1.59)}^2}} }}} \right)10\angle – 23.2^\circ \end{array}$

 

Vout = 9.20∠−23.2° V

 

Ques.43. The phase angle in the RC lead circuit is always _______

  1. Negative
  2. Zero
  3. Positive
  4. None of the above

Answer.3. Positive

Explanation:-

An RC lead circuit is a phase shift circuit in which the output voltage leads the input voltage by a specified amount.

RC lead

 

Ques.44. In the series RC lead circuit, the output is taken across_______

  1. Resistor
  2. Capacitor
  3. Both 1 & 2
  4. None of the above

Answer.1. Resistor

Explanation:-

When the output of a series RC circuit is taken across the resistor rather than across the capacitor, it becomes a lead circuit.

RC lead circuit

 

Ques.45. An ______ circuit is a phase shift circuit in which the output voltage leads the input voltage by a specified amount.

  1. RC Lag
  2. RC Lead
  3. Voltage Lead
  4. Current Lead

Answer.2. RC Lead

Explanation:-

An RC lead circuit is a phase shift circuit in which the output voltage leads the input voltage by a specified amount. When the output of a series RC circuit is taken across the resistor rather than across the capacitor, it becomes a lead circuit.

RC.img .9

Phase Difference Between Input and Output:-  In a series RC circuit, the current leads to the input voltage. As we know that the resistor voltage is in phase with the current. Since the output voltage is taken across the resistor, the output leads the input, as shown in the phasor diagram.

 

Ques.46. The phase angle of an RC lead circuit is designated as

  1. −tan−1(XC/R)
  2. tan−1(XC/R)
  3. −tan−1(R/XC)
  4. tan−1(R/XC)

Answer.2. tan−1(XC/R)

Explanation:-

As in the lag circuit, the amount of phase difference between the input and output and the magnitude of the output voltage in the lead circuit is dependent on the relative values of the resistance and the capacitive reactance.

When the input voltage is assigned a reference angle of 0° the angle of the output voltage is the same as θ (the angle between total current and applied voltage) because the resistor voltage (output) and the current are in phase with each other. Therefore, since φ= θ in this case, the expression is

φ = tan−1(XC/R)

This angle is positive because the output leads the input.

 

Ques.47. The magnitude of the output voltage of RC phase lead circuit is

(1) ${V_{out}} = \left( {\frac{R}{{\sqrt {{R^2} + {X^2}_C} }}} \right)$

 

(2) ${V_{out}} = \left( {\frac{{{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

 

(3) ${V_{out}} = \left( {\frac{R}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

 

(4) ${V_{out}} = \left( {\frac{{2{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

 

Answer.3. ${V_{out}} = \left( {\frac{R}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

Explanation:-

The magnitude of the Output Voltage:-  Since the output voltage of an RC lead circuit is taken across the resistor, the magnitude can be calculated using either Ohm’s law (Vout = IR) or the voltage-divider formula.

${V_{out}} = \left( {\frac{R}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}$

The phasor expression for the output voltage of an RC lead circuit is

Vout = Vout ∠φ

 

Ques.48. Calculate the output phase angle in the RC lead circuit when the resistance is 220 Ω and capacitive reactance is 150 Ω?

  1. −3.43°
  2. 3.43°
  3. −34.3°
  4. 34.3°

Answer.4. 34.3°

Explanation:-

The phase of an RC lead circuit is

φ = tan−1(XC/R)

= tan−1(150/220)

φ = 34.3°

The output leads the input by 71.6°.

 

Ques.49. Calculate the output phase angle in the RC lead circuit when the resistance is 1 kΩ, capacitance is 0.22µF and the frequency is 500 Hz?

  1. 5.54°
  2. 55.4°
  3. −5.54°
  4. −55.4°

Answer.2. 55.4°

Explanation:-

Capacitive reactance

XC = 1/2πfC

XC = 1/6.28 × 500 × 0.22

XC = 1.45 kΩ

The phase angle of an RC lead circuit is

φ = tan−1(XC/R)

= tan−1(1.45/1)

φ = 55.4°

The output lags the input by 55.4°

 

Ques.50. What will be the magnitude of output voltage in the RC lead circuit when the resistance is 1 kΩ, capacitance is 0.22µF,  frequency is 500 Hz and the input voltage has an RMS value of 10 V?

  1. 5.68∠55.4° V
  2. 568∠-55.4° V
  3. 5.68∠−55.4° V
  4. 568∠55.4° V

Answer.1. 5.68∠55.4° V

Explanation:-

Capacitive reactance

XC = 1/2πfC

XC = 1/6.28 × 500 × 0.22

XC = 1.45 kΩ

The phase angle of an RC lead circuit is

φ = tan−1(XC/R)

= tan−1(1.45/1)

φ = 55.4°

The output voltage in phasor form is

Vout = Vout ∠φ

$\begin{array}{*{20}{l}}{{V_{out}} = \left( {\frac{R}{{\sqrt {{R^2} + {X^2}_C} }}} \right){V_{in}}\angle \phi }\\{}\\{}\\\begin{array}{l} = \left( {\frac{1}{{\sqrt {{{(1)}^2} + {{(1.45)}^2}} }}} \right)10\angle {55.4^^\circ }\\\\= \left( {\frac{1}{{1.76}}} \right)10\angle {55.4^^\circ }\end{array}\end{array}$

 

Vout = 5.68∠55.4° V

 

Scroll to Top