# Important MCQ of RC Circuit with Explanation – 2021 | RC Circuit MCQ

Ques.1. A positive angle of 60° is equivalent to a negative angle of

1. −300°
2. 330°
3. 240°
4. 360°

We subtract 360 from the angle to get its corresponding negative angle.

60° = 60 – 360 = − 300°

Ques.2. When the frequency of the source voltage decreases, the impedance of an RC circuit

1. Remain Same
2. Decrease
3. Increase
4. None of the above

As we know, capacitive reactance varies inversely with frequency.

The impedance of the RC circuit is given as

$Z = \sqrt {{R^2} + {X^2}_C}$

When XC increases, the entire term under the square root sign increases, and thus the magnitude of the total impedance also increases; and when XC decreases, the magnitude of the total impedance also decreases. Therefore, in a series RC circuit, Z is inversely dependent on frequency.

As the frequency is increased, XC decreases; so less voltage is dropped across the capacitor. Also, Z decreases as XC decreases, causing the current to increase. An increase in the current causes more voltage across R. As the frequency is decreased, increases, XC  so more voltage is dropped across the capacitor. Also, Z increases as XC increases, causing the current to decrease. A decrease in the current causes less voltage across R.

As the frequency increases, the voltage across Z remains constant because the source voltage is constant. Also, the voltage across C decreases. The increasing current indicates that Z is decreasing. It does so because of the inverse relationship stated in Ohm’s law i.e Z = Vs/I.  The increasing current also indicates that XC is decreasing (XC = Vc/I). The decrease in Vc corresponds to the decrease in XC.

Since XC is the factor that introduces the phase angle in a series RC circuit, a change in XC produces a change in the phase angle. As the frequency is increased, becomes smaller, and thus the phase angle decreases. As the frequency is decreased, XC becomes larger, and thus the phase angle increases.

Ques.3. The complex number 6 + j6 is equivalent to

1.
2. 8.48 ∠90°

The magnitude of the phasor represented by 6 + j6 is

$\begin{array}{l} C = \sqrt {{A^2} + {B^2}} \\ \\ C = \sqrt {{8^2} + {6^2}} \end{array}$

C = 8.48

Since the phasor is in the first quadrant, therefore, the angle is

θ = tan−1(±B/A)

θ = tan−1(6/6) = 45°

θ is the angle relative to the positive real axis. The polar form of 8 + j6 is

C∠θ = 8.48∠45°

Ques.4. In an RC circuit, the impedance is determined by both the ______ and the ________combined.

1. Resistance and Inductive Reactance
2. Capacitive Reactance and Inductive Reactance
3. Resistance and Inductance
4. Resistance and capacitive Reactance

In an RC circuit, the impedance is determined by both the Resistance and the Capacitive reactance combined.

$Z = \sqrt {{R^2} + {X^2}_C}$

Ques.5. The conversion of 10∠30° from polar quantities to rectangular form is

1. 8.66 + j5
2. 8.66 − j5
3. 4.66 + j5
4. 4.66 − j5

The real part of the phasor represented by 10∠30° is

A = C.cosφ

= 10 cos 30° = 10(0.866) = 8.66

The imaginary part j of the phasor is

jB = jC sinφ

= j10 sin 30° = j10(0.5) = j5

The rectangular form of 10∠30° is

A + jB = 8.66 + j5

Ques.6. The RC capacitor circuit the capacitor voltage always leads the current by ______

1. 90°
2. 45°
3. 0° to 90°

When a sinusoidal voltage is applied to a series RC circuit, each resulting voltage drop and the current in the circuit are also sinusoidal and have the same frequency as the applied voltage. The capacitance causes a phase shift between the voltage and current that depends on the relative values of the resistance and the capacitive reactance. As shown in Figure, the resistor voltage VR the capacitor voltage  VC and the current (I) are all sine waves with the frequency of the source. Phase shifts are introduced because of the capacitance. The resistor voltage and current lead to the source voltage, and the capacitor voltage lags the source voltage. The phase angle between the current and the capacitor voltage is always 90°.

The amplitudes and the phase relationships of the voltages and current depend on the resistance and capacitive reactance values. When a circuit is purely resistive, the phase angle between the applied (source) voltage and the total current is zero.

When a circuit is purely capacitive, the phase angle between the applied voltage and the total current is 90° with the current leading the voltage. When there is a combination of both resistance and capacitive reactance in a circuit, the phase angle between the applied voltage and the total current is somewhere between 0° and 90° depending on the relative values of the resistance and the capacitive reactance.

Ques.7. Conversion of 8 + j6 complex numbers from rectangular form to polar form is

1. 20∠36.9°
2. 10∠63.9°
3. 20∠63.9°
4. 10∠36.9°

The magnitude of the phasor represented by 8 + j6 is

$\begin{array}{l} C = \sqrt {{A^2} + {B^2}} \\ \\ C = \sqrt {{8^2} + {6^2}} \end{array}$

C = 10

Since the phasor is in the first quadrant, therefore, the angle is

θ = tan−1(±B/A)

θ = tan−1(6/8) = 36.9°

θ is the angle relative to the positive real axis. The polar form of 8 + j6 is

C∠θ = 10∠36.9°

Ques.8.  The complex plane, the number 24 – j8 is located in

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

$C = \sqrt {{A^2} + {B^2}}$ So from the above figure, it is clear that the number 24 – j8 is located in the fourth quadrant.

Ques.9. The total current in an RC circuit ______  the source voltage.

1. Lag
4. None of the above

In a series RC circuit, the current is the same through both the resistor and the capacitor. Thus, the resistor voltage is in phase with the current, and the capacitor voltage lags the current by 90°. Therefore, there is a phase difference of 90° between the resistor voltage, VR, and the capacitor voltage VC, as shown in the waveform diagram. Ques.10. The complex number 200∠-45° is equivalent to

1. 141 + j141
2. 1.41 + j1.41
3. 1.41 − j1.41
4. 141 − j141

The real part of the phasor represented by 10∠30° is

A = C.cosφ

= 200 cos (−45°) = 200(0.707) = 141

The imaginary part j of the phasor is

jB = jC sinφ

= j200 sin (−45°) = 200(−0.707) = −j141

The rectangular form of 10∠30° is

A + jB = 141 −j141

Ques.11. _______ and ______ are two forms of complex numbers that are used to represent phasor quantities.

1. Rectangular, Polar
2. Polar, Square
3. Square, Rectangular
4. None of the above

Rectangular and Polar Forms

Rectangular and polar are two forms of complex numbers that are used to represent phasor quantities. Each has certain advantages when used in circuit analysis, depending on the particular application. A phasor quantity contains both magnitude and angular position or phase. In this text, italic letters such as V and I are used to represent magnitude only, and boldfaced nonitalic letters such as V and I are used to represent complete phasor quantities.

Ques.12. The addition of complex number 8 + j5 and 2 + j1 is equivalent to

1. 10 − j6
2. 10 + j6
3. 20 − j6
4. 20 + j6

(8 + j5) + (2 + j1)

= (8 + 2) + j(5 + 1)

= 10 + j6

Ques.13. The subtraction of complex number 20 − j10 and 12 + j6 is equivalent to

1. 32 − j10
2. 40 − j10
3. 32 − j4
4. 23 − j4

(20 – j10) + (12 + j6)

= (20 + 12) + j(-10 + 6)

= 32 + j(-4) = 32 − j4

Ques.14. A 2 kΩ resistor and a 0.002 µF capacitor are in series across an ac source. The current in the circuit is 6.50 mA. The true power is

1. 84.5 mW
2. 845 mW
3. 8.45 mW
4. 0.845 mW

Given

Resistance R = 2 kΩ

Current I = 6.50 mA

Due to capacitance or inductance, we get reactive power (Q).

The real power (S) depends on only resistance.

S = VI = I2/R

S = (6.5)2/2000

P = 0.845 W

P = 84.5 W

Ques.15. What will be the phasor expression for the impedance in rectangular form when the 56 Ω Resistance and 100 Ω capacitive Reactance are connected in Series?

1. 56Ω + j100Ω
2. 28Ω + j150Ω
3. 56Ω − j100Ω
4. 28Ω − j150Ω

The impedance is simply the capacitive reactance, and the phase angle is −90° because the capacitance causes the current to lead the voltage by 90°.

Z = R − jXC

Z = 56Ω − j100Ω

Ques.16. What will be the phasor expression for the impedance in polar form when the 56 Ω Resistance and 100 Ω capacitive Reactance are connected in Series?

1. 115 ∠− 60.8°Ω
2. 1.15 ∠− 60.8°Ω
3. 115 ∠− 608°Ω
4. 115 ∠− 6.08°Ω

The impedance in polar form for series RC circuit is

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(56)}^2} + {{(100)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{100}}{{56}}} \right) \end{array}$

Z = 115 ∠− 60.8°Ω

Ques.17. The current in the given Figure is expressed in polar form as I = 0.2∠0°. What will be the source voltage expressed in polar form? 1. 376∠− 57.8°Ω
2. 37.6∠− 60.8°Ω
3. 3.76∠− 57.8°Ω
4. 376∠− 5.78°Ω

The magnitude of the capacitive reactance is

XC = 1/2πfC

XC = 1/2 × 1000 × 0.01

XC = 15.9 kΩ

The total impedance in rectangular form is

Z = R − jXC

Z = 10kΩ − j15.9kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(10)}^2} + {{(15.9)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{15.9}}{{10}}} \right) \end{array}$

Z = 18.8 ∠− 57.8° kΩ

Use Ohm’s law to determine the source voltage

Vs = IZ

= (0.2∠0°)(18.8∠-57.8° kΩ)

= 3.76∠− 57.8° Ω

Ques.18. In a series RC circuit, the Resistance is 2.2 kΩ, capacitance is 0.022µF, Frequency is 1.5 kHz and the source voltage is 10∠0°. Determine the value of current in the series RC circuit.

The magnitude of the capacitive reactance is

XC = 1/2πfC

XC = 1/6.28 × 1.5 × 0.022

XC = 4.82

The total impedance in rectangular form is

Z = R − jXC

Z = 2.2 kΩ − j4.82 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(2.2)}^2} + {{(4.82)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{4.82}}{{2.2}}} \right) \end{array}$

Z = 5.30 ∠− 65.5° kΩ

Use Ohm’s law to determine the source current

I = V/Z

= (10∠0°)/5.30 ∠− 65.5°)

= 1.89∠65.5° mA

Ques.19.When the frequency is increased the capacitive reactance of the circuit is

1. Increased
2. Decreased
3. Remain same
4. Does not depend on the frequency

The magnitude of the capacitive reactance is

XC = 1/2πfC

XC ∝ 1/f

Since the Capacitive reactance is inversely proportional to the frequency, when the frequency increases, the capacitive reactance decreases.

Ques.20. In RC circuit the impedance is ________ Proportional to the Frequency.

1. Directly
2. Indirectly
3. Both 1 and 2
4. None of the above

As we know, capacitive reactance varies inversely with frequency.

The impedance of the RC circuit is given as

$Z = \sqrt {{R^2} + {X^2}_C}$

When XC increases, the entire term under the square root sign increases, and thus the magnitude of the total impedance also increases; and when XC decreases, the magnitude of the total impedance also decreases. Therefore, in a series RC circuit, Z is inversely dependent on frequency.

Ques.21. In a series RC circuit, when the frequency and the resistance are halved, the impedance

1. Doubles
2. Halved
3. One-fourth
4. Can’t be determined

The magnitude of the capacitive reactance is

XC = 1/2πfC

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \\ \\ Z = \sqrt {{R^2} + \frac{1}{{4{\pi ^2}{f^2}_C}}} \end{array}$

If frequency and resistance are halved, we cannot determined the impedance unless the numerical values are given.

Ques.22. A 2 kΩ resistor is in series with a 0.015 µF capacitor across a 15 kHz ac source. What are the magnitude of the total impedance and the phase angle?

1. 707 Ω and ∠ = -19.5°
2. 73.4 Ω and ∠ = -19.5°
3. 2121 Ω and ∠ = -19.5°
4. 734 Ω and ∠ = -38.9°

The magnitude of the capacitive reactance is

XC = 1/2πfC

XC = 1/6.28 × 15 × 0.015

XC = 0.707

The total impedance in rectangular form is

Z = R − jXC

Z = 2 kΩ − j2.22 kΩ

Converting it to polar form

$\begin{array}{*{20}{l}} {Z = \sqrt {{R^2} + {X^2}_C} \angle – {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)}\\ \begin{array}{l} \\ \end{array}\\ {Z = \sqrt {{{(2)}^2} + {{(0.707)}^2}} \angle – {{\tan }^{ – 1}}\left( {\frac{{0.707}}{2}} \right)} \end{array}$

Z = 2121Ω ∠− 19.5° kΩ

Ques.23. What is the angular difference between +j2 and -j2?

1. 180°
2. 90°
3. 30°
4. 360°

If you multiply the positive real value of  +2 by j, the result is +j2. This multiplication has effectively moved the +2 through an angle to the axis.

Similarly, multiplying +2 by −j rotates it −90° to the  −j  axis. Thus, j is considered a rotational operator.

Mathematically, the j operator has a value of √−1.  If +j2 is multiplied by j, you get

j22 = (√−1) (√−1)(2) = (−1)(2) =−2

This calculation effectively places the value on the negative real axis. Therefore, multiplying a positive real number by j2 converts it to a negative real number, which, in effect, is a rotation of 180° on the complex plane. Ques.24.When the frequency of the voltage applied to a series RC circuit is increase, the impedance

1. Increase
2. Decrease
3. Remain same
4. None of the above

As we know, capacitive reactance varies inversely with frequency.

The impedance of the RC circuit is given as

$Z = \sqrt {{R^2} + {X^2}_C}$

When XC increases, the entire term under the square root sign increases, and thus the magnitude of the total impedance also increases; and when XC decreases, the magnitude of the total impedance also decreases. Therefore, in a series RC circuit, Z is inversely dependent on frequency.

As the frequency is increased, XC decreases; so less voltage is dropped across the capacitor. Also, Z decreases as XC decreases, causing the current to increase. An increase in the current causes more voltage across R. As the frequency is decreased, increases, XC  so more voltage is dropped across the capacitor. Also, Z increases as XC increases, causing the current to decrease. A decrease in the current causes less voltage across R.

As the frequency increases, the voltage across Z remains constant because the source voltage is constant. Also, the voltage across C decreases. The increasing current indicates that Z is decreasing. It does so because of the inverse relationship stated in Ohm’s law i.e Z = Vs/I.  The increasing current also indicates that XC is decreasing (XC = Vc/I). The decrease in Vc corresponds to the decrease in XC.

Since XC is the factor that introduces the phase angle in a series RC circuit, a change in XC produces a change in the phase angle. As the frequency is increased, becomes smaller, and thus the phase angle decreases. As the frequency is decreased, XC becomes larger, and thus the phase angle increases.

Ques.25. In series RC circuit the _____ leading _________

1. Voltage, Current
2. Power, Current
3. Current, Power
4. Current, Voltage

When a sinusoidal voltage is applied to a series RC circuit, each resulting voltage drop and the current in the circuit are also sinusoidal and have the same frequency as the applied voltage. The capacitance causes a phase shift between the voltage and current that depends on the relative values of the resistance and the capacitive reactance. When a circuit is purely capacitive, the phase angle between the applied voltage and the total current is 90° with the current leading the voltage. When there is a combination of both resistance and capacitive reactance in a circuit, the phase angle between the applied voltage and the total current is somewhere between 0° and 90° depending on the relative values of the resistance and the capacitive reactance.

Ques.26. The circuit phase angle is the angle between the total _______ and the applied _______.

1. Voltage, current
2. Current, voltage
3. Current, Power
4. None of the above
• The phase angle is the angle between the total current and source voltage. It may be positive or negative depending on whether the source voltage leads or lags the circuit current.
• When a circuit is purely resistive, the phase angle between the applied (source) voltage and the total current is zero.
• When a circuit is purely capacitive, the phase angle between the applied voltage and the total current is with the current leading the voltage.
• When there is a combination of both resistance and capacitive reactance in a circuit, the phase angle between the applied voltage and the total current is somewhere between 0 and 90° depending on the relative values of the resistance and the capacitive reactance.

Ques.27. A capacitor with 150Ω of capacitive reactance is across an ac source. The impedance, expressed in polar form, is

1. Z = 150
2. Z = 150 ∠-90°
3. Z = 150 ∠90°
4. Z = 150 ∠180°

The total impedance in rectangular form is

Z = R − jXC

As R which is not given, it’s negligible

Z = − j2.22 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(150)}^2}} \angle – {\tan ^{ – 1}}\left( {150} \right) \end{array}$

Z = 150 ∠-90°

Ques.28. A 12 kΩ resistor is in series with a 0.02 µF capacitor across a 1.2 kHz ac source. If the current is expressed in polar form as I = 0.3 ∠0° mA, what is the source voltage expressed in polar form?

1. 45.6 V
2. 411.3 V
3. 0.411 V
4. 4.11 V

The magnitude of the capacitive reactance is

XC = 1/2πfC

XC = 1/6.28 × 1.2 × 0.02

XC = 6.67

The total impedance in rectangular form is

Z = R − jXC

Z = 12 kΩ − j6.67 kΩ

Converting it to polar form

$\begin{array}{l} Z = \sqrt {{R^2} + {X^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)\\ \\ Z = \sqrt {{{(2.2)}^2} + {{(4.82)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{4.82}}{{2.2}}} \right) \end{array}$

Z =13.72 ∠− 28.81° kΩ

Use Ohm’s law to determine the source voltage

Vs = IZ

= (0.3∠0°)(13.72∠-28.81° kΩ)

= 4.116∠− 28.81° Ω

Ques.29. An imaginary number exists only in the mind of the mathematician

1. True
2. False
3. It May be true or false
4. None of the above

The product of two positive numbers, as well as two negative numbers, is positive, how can we find the square root of a negative number. Or, can we find a number such that, when it is multiplied by itself gives a negative number. Obviously not. But there arise situations where we have to find the square root of negative numbers.

Solving for x from the equation x2 +25 = 0 is such an instance. For enabling these operations, is taken and denoted by I, the first letter of the word imaginary (In fact it is the Greek letter iota). Using this notation, we can write the square root of negative numbers.

An imaginary number that when squared gives a negative result ie i(iota). Imaginary numbers do exist. Despite their name, they are not really imaginary at all.

Ques.30. For a certain load, the true power is 150 W and the reactive power is 125 VAR. The apparent power is

1. 275 W
2. 195.2 W
3. 19.52 W
4. 25 W

True power P = 150W

Reactive power Pr = 125W

Apparent Power PA

$\begin{array}{l} {P_A} = \sqrt {{P^2} + {P^2}_R} \\ \\ {P_A} = \sqrt {{{(150)}^2} + {{(125)}^2}} \end{array}$

PA = 195.2 W

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

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

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

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

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

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

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

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

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)

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

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

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°

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°.

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

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. 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

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

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

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

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. 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)

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

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°

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°

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

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

Ques.51. The expression of total impedance develop in the parallel RC circuit is

(1) ${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)}$

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

(3)${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle {{\tan }^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)}$

(4) ${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {{\tan }^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)}$

# Parallel RC Circuit Impedance

Since there are only two circuit components, R and C, the total impedance can be expressed by the product-over-sum rule. ${Z = \frac{{\left( {R\angle {0^^\circ }} \right)\left( {{X_C}\angle – {{90}^^\circ }} \right)}}{{R – j{X_C}}}}$

By multiplying the magnitudes, adding the angles in the numerator, and converting the denominator to polar form, we get

$Z = \frac{{\left( {R{X_C}} \right)\angle \left( {{0^^\circ } – {{90}^^\circ }} \right)}}{{\sqrt {{R^2} + {X^2}_C} \angle – {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)}}$

Now, by dividing the magnitude expression in the numerator by that in the denominator, and by subtracting the angle in the denominator from that in the numerator

$Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle \left( { – {{90}^^\circ } + {{\tan }^{ – 1}}\left( {\frac{{{X_c}}}{R}} \right)} \right)$

Equivalently, this expression can be written as

${Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {{\tan }^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)}$

Ques.52. For the RC Parallel determine the magnitude and phase angle of the total impedance when the resistance is 100Ω, capacitive reactance is 50Ω and the source voltage is 5V.

1. 4.47 ∠−63.4°Ω
2. 44.7 ∠−63.4°Ω
3. 4.47 −6.34°Ω
4. 40.7 −60.3°Ω

For the Parallel RC circuit, the total Impedance is

$\begin{array}{l} Z = \left( {\frac{{R{X_C}}}{{\sqrt {{R^2} + {X^2}_C} }}} \right)\angle – {\tan ^{ – 1}}\left( {\frac{R}{{{X_c}}}} \right)\\ \\ \\ = \frac{{(100)(50)}}{{\sqrt {{{(100)}^2} + {{(50)}^2}} }}\angle – {\tan ^{ – 1}}\left( {\frac{{100}}{{50}}} \right) \end{array}$

= 44.7 ∠−63.4°Ω

Ques.53. For the RC Parallel circuit shown in the figure, determine the magnitude and phase angle of the total impedance. 1. 8.94∠−26.6°Ω
2. 894∠26.6°Ω
3. 894∠−26.6°Ω
4. 8.94∠26.6°

From the given figure

Resistance = 1 kΩ

Inductance = 2 kΩ

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

Z = 894∠−26.6°Ω

Ques.54. For parallel RC circuits, the phasor expression for Conductance (G) is

1. G∠90°
2. G∠180°
3. G∠−90°
4. G∠0°

The conductance, G, is the reciprocal of resistance. The phasor expression for conductance is expressed as

G = 1/R∠0° = G∠0°

Ques.55. For parallel RC circuits, the phasor expression for inductive susceptance (BC) is

1. jBC
2. −jBC
3. 2jBL
4. None of the above

Capacitive susceptance (BC) is the reciprocal of capacitive reactance. The phasor expression for capacitive susceptance is

BC = 1/XC∠−90°

BC = ∠90° = jBC

Ques.56. For parallel RC circuits, the phasor expression for Admittance (Y) is

1. Y∠θ
2. Y∠−θ
3. Y∠2θ
4. Y∠±θ

Admittance (Y) is the reciprocal of impedance. The phasor expression for admittance is

Y = 1/Z∠ ±θ

Y = ∠ ±θ

Ques.57. For parallel RC circuits, the phasor expression for total Admittance (Y) of the circuit is

(1) $Y = \sqrt {{G^2} – {B^2}_C}$

(2) $Y = \sqrt {{G^2} + {B^2}_C}$

(3) $Y = \sqrt {{G^2} + 2{B^2}_C}$

(4) None of the above

In working with parallel circuits, it is often easier to use conductance (G), capacitive susceptance, and admittance (Y) rather than resistance (R), capacitive reactance and impedance (Z).

In a parallel RC circuit, as shown in Figure a, the total admittance is simply the phasor sum of the conductance and the capacitive susceptance. Y = G + jBC

$Y = \sqrt {{G^2} + {B^2}_C}$

Ques.58. Determine the total admittance of the RC parallel circuit when the resistance is 330Ω, capacitance is 0.22μF  and the frequency is 1 kHz.

1. 30.3 + j1.38
2. 3.03 − j1.38
3. 3.03 + j1.38
4. 30.3 − j1.38

Resistance R = 330Ω

Conductance = 1/R = 1/330 = 3.03 mS.

Now Capacitive reactance

Xc = 1/2πfC

XC = 1/2π × 1000 × 0.22 = 723 kΩ

Capacitive Suspectance

BC = 1/XC = 1/723 = 1.38 mS.

Yt = G + JBC

Yt = 3.03 + j1.38

Ques.59. Determine the total impedance of the RC parallel circuit in polar form when the resistance is 330Ω, capacitance is 0.22μF  and the frequency is 1 kHz.

1. 300∠24.5° mS
2. 30.0∠24.5° mS
3. 300∠−24.5° mS
4. 30∠−24.5° mS

Resistance R = 330Ω

Conductance = 1/R = 1/330 = 3.03 mS.

Now Capacitive reactance

Xc = 1/2πfC

XC = 1/2π × 1000 × 0.22 = 723 kΩ

Capacitive Suspectance

BC = 1/XC = 1/723 = 1.38 mS.

Yt = G + JBC

Yt = 3.03 + j1.38

Converting to polar form

$\begin{array}{l} {Y_t} = \sqrt {{G^2} + {B^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_c}}}{G}} \right)\\ \\ = \sqrt {{{(3.03)}^2} + {{(1.38)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{1.38}}{{3.03}}} \right) \end{array}$

Now total impedance

Zt = 1/Yt

Zt = 1/(3.03 + j1.38)

Zt = 300∠−24.5° mS

Ques.60. Determine the total current of parallel RC circuit  as shown in the figure below 1. 50∠°−24.5° mA
2. 50∠24.5° mA
3. 500∠−24.5° mA
4. 5∠24.5° mA

Resistance R = 330Ω

Conductance = 1/R = 1/2.2 = 455 µS.

Now Capacitive reactance

Xc = 1/2πfC

XC = 1/2π × 1.5 × 0.022 = 4.82 kΩ

Capacitive Suspectance

BC = 1/XC = 1/4.82 = 207 µS.

Yt = G + JBC

Yt = 455 + j207

Converting to polar form

$\begin{array}{l} {Y_t} = \sqrt {{G^2} + {B^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_c}}}{G}} \right)\\ \\ = \sqrt {{{(455)}^2} + {{(207)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{207}}{{455}}} \right) \end{array}$

Yt = 500∠24.5° mS

Use Ohm’s law to determine the total current.

Itot = Vs.Yt

Itot = (10∠0°)(500∠24.5°)

Itot = 5∠24.5° mA

Ques.61. The total current in the Parallel RC circuit is expressed as

1. It = IR + IjC
2. It = IR − IjC
3. IR = It + IjC
4. None of the above

# Phase Relationships of Currents and Voltages for the parallel RC circuit

Figure (a) shows all the currents in a basic parallel RL circuit. The total current It divides at the junction into the two branch currents, IR and IC. The applied voltage Vs appear across both the resistive and the inductive branches, so Vs, Vr, and VC are all in phase and of the same magnitude. The current through the resistor is in phase with the voltage. The current through the capacitor leads the voltage and the resistor current by 90°. By Kirchhoff’s current law, the total current is the phasor sum of the two branch currents, as shown by the phasor diagram in Figure (b). The total current is expressed as

It = IR + IjC

This equation can be expressed in polar form as

${I_t} = \sqrt {I_R^2 + I_C^2} \angle {\tan ^{ – 1}}\left( {\frac{{{I_C}}}{{{I_R}}}} \right)$

Ques.62. Find the total current in the Parallel RC circuit shown in the figure below 1. 54.5 A − j80 A
2. 80 mA − 54.5 mA
3. 54.5 mA + j80 mA
4. 54.5 mA − j80 mA

From the given figure the resistor current, the inductor current, and the total current are

IR = Vs/R = 12∠0°/220∠0° = 54.5∠0° mA

IL = Vs/XC = 12∠0°/150∠90° = 80∠90° mA

It = IR + IjC

It = 54.5 mA + j80 mA

Ques.63. For conversion of a parallel RC circuit to Series RC circuit, the magnitude of ______ and _______ should be identical.

1. Resistance, Impedance
2. Impedance, Reactance
3. Reactance, Conductance
4. Impedance, Phase angle

For every parallel RC circuit, there is an equivalent series RC circuit for a given frequency. Two circuits are considered equivalent when they both present an equal impedance at their terminals; that is, the magnitude of impedance and the phase angle are identical. To obtain the equivalent series circuit for a given parallel RC circuit, we need to find the impedance and phase angle of the parallel circuit. Then use the values of Z and θ to construct an impedance triangle, shown in Figure. The vertical and horizontal sides of the triangle represent the equivalent series resistance and capacitive reactance as indicated. These values can be found using the following trigonometric relationships:

Req = Zcosθ

XCeq = Zsinθ

Ques.64. Convert the parallel circuit into a series form when the resistance is 18 kΩ and capacitance reactance is 27 kΩ.

1. 1.25 kΩ − j83.1 kΩ
2. 12.5 kΩ − j8.31 kΩ
3. 1.25 kΩ + j83.1 kΩ
4. 12.5 kΩ + j8.31 kΩ

Resistance R = 330Ω

Conductance = 1/R = 1/18 = 55.6 µS.

Capacitive Suspectance

BC = 1/XC = 1/27 = 37.0 µS.

Yt = G + JBC

Yt = 55.6 + j37.0

Converting to polar form

$\begin{array}{l} {Y_t} = \sqrt {{G^2} + {B^2}_C} \angle – {\tan ^{ – 1}}\left( {\frac{{{B_c}}}{G}} \right)\\ \\ = \sqrt {{{(55.6)}^2} + {{(37)}^2}} \angle – {\tan ^{ – 1}}\left( {\frac{{37}}{{55.6}}} \right) \end{array}$

Yt = 66.8∠33.6° mS

Now total impedance

Zt = 1/Yt

Zt = 1/(3.03 + j1.38)

Zt = 15.0∠−33.6° kΩ

Converting to rectangular form yields

Ztot = ZCosθ − jZsinθ = Req − jXC(eq)

= 15.Cos(−33.6°) − j15.sin(−33.6°)

Ztot = 12.5 kΩ − j8.31 kΩ

Ques.65. When the value of the resistor in a series RC circuit is greater than the capacitive Reactance then

1. More energy is stored
2. More energy is dissipated
3. No energy loss
4. None of the above

When the resistance in a series RC circuit is greater than the capacitive reactance, more of the total energy delivered by the source is converted to heat by the resistance than is stored by the capacitor. Likewise, when the reactance is greater than the resistance, more of the total energy is stored and returned than is converted to heat.

Ques.66. In a series RC circuit, 12 V(RMS) is measured across the resistor and 15 V(RMS) is measured across the capacitor. The RMS source voltage is

1. 3 V
2. 27 V
3. 19.2 V
4. 1.9 V

In series RC circuit the voltage is additive

$\begin{array}{*{20}{l}} {{E_{rms}} = \sqrt {{V^2}_R + {V^2}_C} }\\ {}\\ {{E_{rms}} = \sqrt {{{(12)}^2} + {{(15)}^2}} } \end{array}$

ERMS = √369

ERMS =19.2 V

Ques.67. An ac circuit consists of a resistor and a capacitor. To increase the phase angle above 45°, the following condition must exist:

1. R < XC
2. R = XC
3. R > XC
4. None of the above

The phase angle of an RC lead circuit is

φ = tan−1(XC/R)

To increase the phase angle we need to increase the value of capacitive reactance or decrease the value of resistance since capacitive reactance is directly proportional to the capacitive reactance.

XC > R

Ques.68. A resistor and a capacitor are in series across a 20 V ac source. Circuit impedance is 4.33 kΩ. The current flow in the circuit is

1. 460 mA
2. 92 mA
3. 4.6 mA
4. 460 mA

Given

Z = 4.33 kΩ

V = 20 V

The total current in the circuit I

I = V/Z = 20/1.33

I = 4.6 mA

Ques.69. As the phase angle between applied voltage and total current ______, the power factor _____

1. Increase, Decrease
2. Increase, Increase
3. Decrease, Decrease
4. Increase, Remain same

The term Cosθ is called the power factor and is stated as

PF = Cosθ

As the phase angle between applied voltage and total current increases, the power factor decreases, indicating an increasingly reactive circuit. The smaller the power factor, the smaller the power dissipation.

Ques.70. A 47 Ω resistor and a capacitor with 150 Ω of capacitive reactance are in series across an ac source. The impedance, expressed in rectangular form, is

1. Z = 47 Ω − j150 Ω
2. Z = 197 Ω

Given

Resistance R = 47 Ω

Capacitive reactance XC = 150 Ω

The total impedance in rectangular form for series RC circuit is

Z = R − jXC

Z = 47Ω − j150 Ω

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