RC Circuit Question and Answers – 2022

Ques.31. A 47 Ω resistor and a capacitor with a capacitive reactance of 120 Ω are in series across an ac source. What is the circuit impedance, Z?

  1. 129 Ω
  2. 12.9 Ω
  3. 167 Ω
  4. 1.29 Ω

Answer.1. 129 Ω

Explanation:-

The total impedance in rectangular form is

Z = R − jXC

Z = 47 Ω − j120 Ω

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \\ \\ Z = \sqrt {{{(47)}^2} + {{(120)}^2}} \end{array}$

Z = 128.87 Ω ≅ 129Ω

 

Ques.32. In the complex plane, the number 4 + j3 is located in the

  1. Fourth quadrant
  2. Second quadrant
  3. Third quadrant
  4. First quadrant

Answer.4. First quadrant

Explanation:-

A phasor can be visualized as forming a right triangle in the complex plane, as indicated in Figure, for each quadrant location. The horizontal side of the triangle is the real value, A, and the vertical side is the j value, B. The hypotenuse of the triangle is the length of the phasor, C, representing the magnitude, and can be expressed, using the Pythagorean theorem, as

$C = \sqrt {{A^2} + {B^2}} $

RC.img .1

So from the above figure, it is clear that the number 4 + j3 is located in the first quadrant.

 

Ques.33. In the complex plane, the number −4 − j3 is located in the

  1. Fourth quadrant
  2. Second quadrant
  3. Third quadrant
  4. First quadrant

Answer.3. Third quadrant

Explanation:-

A phasor can be visualized as forming a right triangle in the complex plane, as indicated in Figure, for each quadrant location. The horizontal side of the triangle is the real value, A, and the vertical side is the j value, B. The hypotenuse of the triangle is the length of the phasor, C, representing the magnitude, and can be expressed, using the Pythagorean theorem, as

$C = \sqrt {{A^2} + {B^2}} $

RC.img .1

So from the above figure, it is clear that the number −4 − j3 is located in the Third quadrant.

 

Ques.34. A 1 kΩ resistor and a capacitor with 0.01µf  are in series across an ac source with 10 kHz frequency. What is the circuit impedance, Z?

  1. 188 ∠−57.8
  2. 1.88 ∠−57.8
  3. 18.8 ∠−57.8
  4. 0.188∠−57.8

Answer.2. 1.88 ∠−57.8

Explanation:-

Given

f = 10 kHz

Capacitive reactance

XC = 1/2πfC

XC = 1/6.28 × 10 × 0.01

XC = 1.59

The total impedance in rectangular form is

Z = R − jXC

Z = 1 kΩ − j1.59 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(1)}^2} + {{(1.59)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{1.59}}{1}} \right) \end{array}$

Z = 1.88 ∠−57.8

 

Ques.35. An _____ circuit is a phase shift circuit in which the output voltage lags the input voltage by a specified amount.

  1. Current lag
  2. Voltage Lag
  3. RC lag
  4. Power lag

Answer.3. RC lag

Explanation:-

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

RC.img .7

The figure shows a series RC circuit with the output voltage taken across the capacitor. The source voltage is the input, Vin.  As you know θ is the phase angle between the current and the input voltage, is also the phase angle between the resistor voltage and the input voltage because VR and I are in phase with each other.

 

Ques.36. The polar form of an RC lag circuit is given as

  1. Vout = IXC∠(−90° + θ)
  2. Vout = IXC∠(−90° − θ)
  3. Vout = IXC∠(−180° − θ)
  4. Vout = IXC∠(−180° + θ)

Answer.1. Vout = IXC∠(−90° + θ)

Explanation:-

Phase Difference Between Input and Output: As we already know, θ is the phase angle between I and Vin.

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)(XC∠−90°)

Vout = IXC∠(−90° + θ)

The above equation states that the output voltage is at an angle of  (−90° + θ)with respect to the input voltage.

 

Ques.37. The phase angle of an RC lag 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.3. −tan−1(R/XC)

Explanation:-

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)(XC∠−90°)

Vout = IXC∠(−90° + θ)

The angle between Vout and Vin is called as phase angle(phi) φ and it is developed as follows

φ = −90° + tan−1(XC/R)

Equivalently, this angle can be expressed as

φ = −tan−1(R/XC)

 

Ques.38. The phase angle in the RC lag circuit is always _______

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

Answer.1. Negative

Explanation:-

Since in the RC lag circuit the output voltage lags the input voltage therefore the phase angle is always negative.

RC.img .8

 

Ques.39. What will be the total phase lag from input and output in lag circuit when the resistance is 15 kΩ and capacitive reactance is 5 kΩ?

  1. 71.6°
  2. 7.16°
  3. −7.16°
  4. −71.6°

Answer.4. −71.6°

Explanation:-

The phase of an RC lag circuit is

φ = −tan−1(R/XC)

= −tan−1(15/5)

φ = -71.6°

The output lags the input by 71.6°.

 

Ques.40. What will be the total phase lag from input and output in RC lag circuit when the resistance is 680 kΩ,  capacitance is 0.1µF and the frequency is 1 kHz?

  1. 23.2°
  2. −23.2°
  3. 2.32°
  4. −2.32°

Answer.2. −23.2°

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 lags the input by 23.2°.

Scroll to Top