Parallel RC Circuit MCQ Quiz - Objective Question with Answer for Parallel RC Circuit - Download Free PDF

Explanation:

Parallel Resistance (R) and Capacitor (C) Circuit

Definition: A parallel resistance and capacitor circuit is an electrical circuit configuration where a resistor (R) and a capacitor (C) are connected in parallel to a common voltage source. Such circuits are widely used in AC (alternating current) applications to analyze impedance, phase angles, and time constants.

Working Principle: In an AC circuit, a capacitor introduces a phase shift because the current leads the voltage by 90 degrees, while the resistor does not introduce any phase shift (current and voltage are in phase). The combination of these two components results in a net phase angle (ϕ) between the total current and the source voltage, which is dependent on the relative values of resistance (R) and capacitive reactance (X_c).

Phase Angle (ϕ): The phase angle (ϕ) in a parallel R-C circuit is determined by the relationship between the current through the capacitor and the current through the resistor. Mathematically, the phase angle is given by:

tan(ϕ) = I_C / I_R

Where:

• I_C = Current through the capacitor
• I_R = Current through the resistor

Since the current through the capacitor is given by I_C = V / X_C and the current through the resistor is I_R = V / R, the phase angle can also be expressed as:

tan(ϕ) = X_C / R

Where:

• X_C = 1 / (2πfC), the capacitive reactance
• R = Resistance

From the above equation, it is evident that the phase angle ϕ depends on the ratio of X_C (capacitive reactance) to R (resistance).

Effect of Increasing Resistance (R):

When the resistance (R) in the parallel R-C circuit is increased:

• The denominator of the term X_C / R increases, resulting in a decrease in the value of tan(ϕ).
• As tan(ϕ) decreases, the phase angle ϕ also decreases since the tangent function is directly related to the angle.

Hence, when resistance (R) increases, the phase angle (ϕ) decreases.

Correct Option:

The correct answer is:

Option 4: Decreases

As explained above, increasing the resistance reduces the phase angle in a parallel R-C circuit due to the inverse relationship between resistance and tan(ϕ).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Increases

This option is incorrect because increasing resistance (R) in the parallel R-C circuit reduces the phase angle (ϕ). The relationship between tan(ϕ) and resistance is inversely proportional, so an increase in R leads to a decrease in tan(ϕ) and consequently a decrease in ϕ.

Option 2: Becomes zero

This option is incorrect because the phase angle (ϕ) does not become zero simply by increasing the resistance. The phase angle becomes zero only in a purely resistive circuit (when the capacitor is removed or its effect is negligible). In a parallel R-C circuit, the phase angle will decrease but will not reach zero as long as the capacitor is present.

Option 3: Remains the same

This option is incorrect because the phase angle (ϕ) is dependent on the resistance (R) and capacitive reactance (X_C). When R changes, the phase angle also changes. Therefore, the phase angle cannot remain the same when resistance is increased.

Conclusion:

In a parallel R-C circuit, the phase angle (ϕ) decreases when the resistance (R) increases. This is due to the inverse relationship between resistance and tan(ϕ), as shown in the mathematical derivation. Understanding the behavior of phase angles in R-C circuits is essential for analyzing AC circuits and their impedance characteristics.

Parallel RC circuit

In a resistor, the current is in the same phase as the voltage.

In a capacitor, the current leads voltage by 90° 

Hence, for a parallel RC circuit, the angle of current (I) with respect to voltage (V) is between 0° to 90° 

\(I_R={V\over R}\angle0\)

\(I_C={V\over 1/j\omega C}={j\omega CV}\)

\(I_C=\angle90 \space ({\omega CV})\)

The supply current is:

I = IR + j I_C

\(I = {V\over R}+j\omega CV\)

\(\rm I=V\left(\frac{1}{R}+\omega C<90^{\circ}\right)\)

The phasor diagram is drawn as:

1) There is no phase difference between the applied voltage and the voltage across R and C in parallel.

2) The current through the resistive branch is in phase with the applied signal.

3) But the current through the capacitive branch leads its voltage V_c by 90 degrees.

The phasor diagram is drawn as:

1) There is no phase difference between the applied voltage and the voltage across R and C in parallel.

2) The current through the resistive branch is in phase with the applied signal.

3) But the current through the capacitive branch leads its voltage Vc by 90 degrees.

At t ≤ 0, V_c (0) = 0 V

\({V_c}\left( \infty \right) = 5\left( {\frac{3}{{2 + 3}}} \right) = 3V\)

Time constant = τ = R_eq C_eq

R_eq = 2kΩ || 3 kΩ = 1.2 kΩ

C_eq = 5 nF

τ = RC = 6 μsec

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + \left[ {{V_c}\left( 0 \right) - {V_c}\left( \infty \right)} \right]{e^{ - \frac{t}{\tau }}}\)

\(= 3 + \left[ {0 - 3} \right]{e^{ - \frac{t}{{6\mu }}}}\)

\(= 3\left[ {1 - {e^{ - \frac{t}{{6\mu }}}}} \right]V\)

At t = 6 μsec

⇒ V_c(t) = 3 [1 - e^-1] = 1.896 V.

Parallel RC circuit

In a resistor, the current is in the same phase as the voltage.

In a capacitor, the current leads voltage by 90° 

Hence, for a parallel RC circuit, the angle of current (I) with respect to voltage (V) is between 0° to 90° 

\(I_R={V\over R}\angle0\)

\(I_C={V\over 1/j\omega C}={j\omega CV}\)

\(I_C=\angle90 \space ({\omega CV})\)

The supply current is:

I = IR + j I_C

\(I = {V\over R}+j\omega CV\)

\(\rm I=V\left(\frac{1}{R}+\omega C<90^{\circ}\right)\)

The phasor diagram is drawn as:

1) There is no phase difference between the applied voltage and the voltage across R and C in parallel.

2) The current through the resistive branch is in phase with the applied signal.

3) But the current through the capacitive branch leads its voltage Vc by 90 degrees.

At t ≤ 0, V_c (0) = 0 V

\({V_c}\left( \infty \right) = 5\left( {\frac{3}{{2 + 3}}} \right) = 3V\)

Time constant = τ = R_eq C_eq

R_eq = 2kΩ || 3 kΩ = 1.2 kΩ

C_eq = 5 nF

τ = RC = 6 μsec

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + \left[ {{V_c}\left( 0 \right) - {V_c}\left( \infty \right)} \right]{e^{ - \frac{t}{\tau }}}\)

\(= 3 + \left[ {0 - 3} \right]{e^{ - \frac{t}{{6\mu }}}}\)

\(= 3\left[ {1 - {e^{ - \frac{t}{{6\mu }}}}} \right]V\)

At t = 6 μsec

⇒ V_c(t) = 3 [1 - e^-1] = 1.896 V.

Explanation:

Parallel Resistance (R) and Capacitor (C) Circuit

Definition: A parallel resistance and capacitor circuit is an electrical circuit configuration where a resistor (R) and a capacitor (C) are connected in parallel to a common voltage source. Such circuits are widely used in AC (alternating current) applications to analyze impedance, phase angles, and time constants.

Working Principle: In an AC circuit, a capacitor introduces a phase shift because the current leads the voltage by 90 degrees, while the resistor does not introduce any phase shift (current and voltage are in phase). The combination of these two components results in a net phase angle (ϕ) between the total current and the source voltage, which is dependent on the relative values of resistance (R) and capacitive reactance (X_c).

Phase Angle (ϕ): The phase angle (ϕ) in a parallel R-C circuit is determined by the relationship between the current through the capacitor and the current through the resistor. Mathematically, the phase angle is given by:

tan(ϕ) = I_C / I_R

Where:

• I_C = Current through the capacitor
• I_R = Current through the resistor

Since the current through the capacitor is given by I_C = V / X_C and the current through the resistor is I_R = V / R, the phase angle can also be expressed as:

tan(ϕ) = X_C / R

Where:

• X_C = 1 / (2πfC), the capacitive reactance
• R = Resistance

From the above equation, it is evident that the phase angle ϕ depends on the ratio of X_C (capacitive reactance) to R (resistance).

Effect of Increasing Resistance (R):

When the resistance (R) in the parallel R-C circuit is increased:

• The denominator of the term X_C / R increases, resulting in a decrease in the value of tan(ϕ).
• As tan(ϕ) decreases, the phase angle ϕ also decreases since the tangent function is directly related to the angle.

Hence, when resistance (R) increases, the phase angle (ϕ) decreases.

Correct Option:

The correct answer is:

Option 4: Decreases

As explained above, increasing the resistance reduces the phase angle in a parallel R-C circuit due to the inverse relationship between resistance and tan(ϕ).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Increases

This option is incorrect because increasing resistance (R) in the parallel R-C circuit reduces the phase angle (ϕ). The relationship between tan(ϕ) and resistance is inversely proportional, so an increase in R leads to a decrease in tan(ϕ) and consequently a decrease in ϕ.

Option 2: Becomes zero

This option is incorrect because the phase angle (ϕ) does not become zero simply by increasing the resistance. The phase angle becomes zero only in a purely resistive circuit (when the capacitor is removed or its effect is negligible). In a parallel R-C circuit, the phase angle will decrease but will not reach zero as long as the capacitor is present.

Option 3: Remains the same

This option is incorrect because the phase angle (ϕ) is dependent on the resistance (R) and capacitive reactance (X_C). When R changes, the phase angle also changes. Therefore, the phase angle cannot remain the same when resistance is increased.

Conclusion:

In a parallel R-C circuit, the phase angle (ϕ) decreases when the resistance (R) increases. This is due to the inverse relationship between resistance and tan(ϕ), as shown in the mathematical derivation. Understanding the behavior of phase angles in R-C circuits is essential for analyzing AC circuits and their impedance characteristics.

At t = 0^-, the circuit becomes,

V_C (0^-) = 1 V

At t = ∞, the circuit becomes

\({V_c}\left( \infty \right) = 3\left( {\frac{2}{{1 + 2}}} \right) = 2\;V\)

\({V_C}\left( t \right) = \left[ {{V_C}\left( 0 \right) - {V_C}\left( \infty \right)} \right]{e^{ - \frac{t}{\tau }}} + {V_C}\left( \infty \right)\;\)

Time constant (τ) = RC = (1||2) (10 × 10^-6)

\(= \frac{2}{3} \times 10 \times {10^{ - 6}}\)

\({V_C}\left( t \right) = 2 + \left[ {1 - 2} \right]{e^{ - \frac{t}{{\frac{2}{3} \times {{10}^{ - 5}}}}}}\)

\(= 2 - {e^{ - \left( {\frac{{3t}}{{2 \times {{10}^{ - 5}}}}} \right)}}\)

At t = 5μs,

\(= 2 - {e^{ - \left( {\frac{{3 \times 5 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 5}}}}} \right)}}\)

\(= 2{ -e ^{ - 0.75}} = 1.527\;V.\)

Parallel RC circuit

In a resistor, the current is in the same phase as the voltage.

In a capacitor, the current leads voltage by 90° 

Hence, for a parallel RC circuit, the angle of current (I) with respect to voltage (V) is between 0° to 90° 

\(I_R={V\over R}\angle0\)

\(I_C={V\over 1/j\omega C}={j\omega CV}\)

\(I_C=\angle90 \space ({\omega CV})\)

The supply current is:

I = IR + j I_C

\(I = {V\over R}+j\omega CV\)

\(\rm I=V\left(\frac{1}{R}+\omega C<90^{\circ}\right)\)

The phasor diagram is drawn as:

1) There is no phase difference between the applied voltage and the voltage across R and C in parallel.

2) The current through the resistive branch is in phase with the applied signal.

3) But the current through the capacitive branch leads its voltage Vc by 90 degrees.

Concept:

For an RC circuit supplied with a constant step DC voltage source, the capacitor voltage rises satisfying the following transient equation:

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + [({V_c}\left( 0 \right) - {V_c}\left( \infty \right))]{e^{ - \frac{t}{{\tau}}}}\)

Vc(∞) = Final or Steady-State voltage of the Capacitor

Vc(0) = Initial voltage stored by the Capacitor

RC = τ = Time Constant of the series RC circuit

Application:

Given:

\({I_x}\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {10\;\;\;t < 0}\\ {25\;\;\;t < 0} \end{array}} \right.\)

At t = 0^- circuit is at steady state and I_x(0^-) = 10 mA (capacitor will be open circuited)

i(0^-) = 0 A

\({I_{20k}} = 10\;mA \times \frac{{8\;k{\rm{\Omega }}}}{{8\;k{\rm{\Omega }} + 12\;k{\rm{\Omega }} + 20\;k{\rm{\Omega }}}}\)

I_20k = 2 mA

V_c (0^-) = (20k × 2m) = 40 V

Similarly, at t = ∞:

\({I_{20k}} = 25 \times \frac{8}{{40}} = 5\;mA\)

V_c (∞) = 100 V

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + [({V_c}\left( 0 \right) - {V_c}\left( \infty \right))]{e^{ - \frac{t}{{\tau}}}}\)

τ = R_eq C_eq 

R_eq = 20 k ∥ 20 k = 10 k

τ = 10 × 10³ × 5 × 10^-6 = 5 × 10^-2

V_c(t) = 100 + (40 - 100) e^-20t

\(i\left( t \right) = C \times \frac{{d{V_c}\left( t \right)}}{{dt}}\)

\(i\left( t \right) = 5 \times {10^{ - 6}} \times \left[ {60} \right] \times \left[ {20} \right]{e^{ - 20t}} \)

\(i(t)= 6{e^{ - 20t}}\;mA\)

\(i\left( t \right){I_{t = 25\;msec}} = 6{e^{ - 0.5}}\)

\( = 3.6391\;mA\)

Concept:

In the steady-state capacitor acts like an open circuit because its impedance becomes ∞ , whereas the inductor acts like a short-circuit because its impedance becomes 0.

Calculation:

The previous steady-state diagram is shown below:

V_C (0^-) = 2 V = V_C (0⁺)

I (0^-) = 1 A

At t = 0⁺, circuit is

By applying mesh analysis from the 2 V source to the ground through the 1 Ω and 2 Ω resistance, we get:

2 + 1(1 - I(0⁺)) - 2I(0⁺) = 0

3 = 3 I(0⁺)

⇒ I (0⁺) = 1 A

At t ≤ 0, V_c (0) = 0 V

\({V_c}\left( \infty \right) = 5\left( {\frac{3}{{2 + 3}}} \right) = 3V\)

Time constant = τ = R_eq C_eq

R_eq = 2kΩ || 3 kΩ = 1.2 kΩ

C_eq = 5 nF

τ = RC = 6 μsec

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + \left[ {{V_c}\left( 0 \right) - {V_c}\left( \infty \right)} \right]{e^{ - \frac{t}{\tau }}}\)

\(= 3 + \left[ {0 - 3} \right]{e^{ - \frac{t}{{6\mu }}}}\)

\(= 3\left[ {1 - {e^{ - \frac{t}{{6\mu }}}}} \right]V\)

At t = 6 μsec

⇒ V_c(t) = 3 [1 - e^-1] = 1.896 V.