Sinusoidal Response of Series RC Circuit MCQ Quiz - Objective Question with Answer for Sinusoidal Response of Series RC Circuit - Download Free PDF

Explanation:

Time Constant of an RC Series Circuit

Definition: The time constant of an RC series circuit is a measure of the time it takes for the voltage across the capacitor to charge or discharge to approximately 63.2% of its final value (or decrease to 36.8% of its initial value) when subjected to a step input of voltage. It is an important parameter in analyzing the transient response of RC circuits.

Formula: The time constant (denoted by τ, pronounced "tau") for an RC circuit is given by:

τ = R × C

Where:

• R is the resistance in ohms (Ω).
• C is the capacitance in farads (F).

This formula indicates that the time constant is directly proportional to both the resistance and the capacitance of the circuit. A larger time constant implies a slower charging or discharging process, while a smaller time constant indicates a faster transient response.

Working Principle: In an RC series circuit, when a DC voltage is applied, the capacitor begins to charge through the resistor. The rate of charging is influenced by the resistor's resistance and the capacitor's capacitance. The time constant (τ) quantifies the duration over which the capacitor reaches a significant proportion of its full charge.

The voltage across the capacitor (V_c) at any time t during charging is given by:

V_c(t) = V_max × (1 - e^-t/τ)

Here:

• V_max is the maximum voltage applied to the circuit.
• t is the time elapsed since the application of the voltage.
• e is the base of the natural logarithm (approximately 2.718).

During discharging, the voltage across the capacitor decreases exponentially and is given by:

V_c(t) = V_initial × e^-t/τ

Where V_initial is the initial voltage across the capacitor at the start of the discharge.

Correct Option Analysis:

The correct option is:

Option 4: RC

This option is correct because the time constant τ is the product of the resistance (R) and the capacitance (C) in the circuit. This relationship is fundamental to the analysis of RC circuits and is derived from the governing equations of the circuit.

Additional Information

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

Option 1: R/C

This option is incorrect. The ratio R/C does not represent the time constant of an RC circuit. Instead, the time constant is the product of resistance and capacitance, not their ratio.

Option 2: C/R

This option is also incorrect. The ratio C/R does not have any direct relevance to the time constant of an RC circuit. The time constant is determined by multiplying R and C, not dividing them.

Option 3: 1/RC

This option is incorrect as well. While 1/RC is related to the frequency of an RC circuit in certain contexts (such as the cutoff frequency in filters), it does not represent the time constant. The time constant is given by RC, not its reciprocal.

Conclusion:

The time constant (τ) of an RC series circuit is a fundamental parameter that determines the transient response of the circuit. It is defined as the product of the resistance (R) and capacitance (C) in the circuit. This relationship is crucial for understanding how quickly a capacitor charges or discharges in an RC circuit. The correct option, RC, accurately represents the time constant, while the other options are incorrect representations of this concept.

The correct answer is 1) It increases by nine times.

Here's a quick recap of why:

• Relationship: Power (P) = Voltage (V)² / Resistance (R)
• Constant Resistance: R remains the same.
• Tripled Voltage: If V becomes 3V, then P becomes (3V)² / R = 9V² / R.
• Result: The power increases by a factor of 9.T

Correct answer is option (2): an inductor of 1H.

Explanation:

As per the given graph, the current through the element I(t) is lagging, it implies that the element is inductor.

\(I(t)= \frac{V(t)}{Z_{o}}\\ \ I(t)= \frac{sin(t)}{Z_{o}}\)          

The maximum value of the current   \(I(t)_{max} = \frac{1}{\sqrt{2}}\)

\(|Z_{o}|=\frac{|V(t)|}{|I(t)|}\\ =\frac{1}{\frac{1}{\sqrt{2}}}\\ =\sqrt{2}\)

As per the circuit given,

Z_o = R + jωL

\(|Z_{o}|=\sqrt{R^{2}+(j ω L)^{2}}\)

\(\sqrt{2}=\sqrt{1+ ω^{2}L^{2}}\)

Given, ω = 1 rad/sec

\( \sqrt{2} = \sqrt{1+L^{2}}\)

2 = 1 + L²

L = 1 H                           

Concept of series RLC Circuit:

A series RLC circuit connected to a voltage source is as shown:

Apply Kirchoff's voltage law to the R-L-C series circuit,

\(Ri + L\frac{{di}}{{dt}} + \frac{1}{C}\mathop \smallint \limits_0^t idt = 0\)

Taking Laplace transform of the above equation,

We get,

\(RI(s)+sLI(s)+\frac{I(s)}{Cs}-i(0^-) +[\frac{I(s)}{s}+\frac{i(0^-)}{s}] =0\)

The initial current i (0^-) is 0. i_c(0^-) is the initial charge across the capacitor and equals - CV₀.

∴ \(RI(s)+sLI(s)+\frac{I(s)}{Cs} +[\frac{I(s)}{Cs}+\frac{V_0}{s}] =0\) 

On simplifying the equation, we get

\(I\left( s \right) = \frac{{{V_0}/L}}{{{s^2} + \frac{R}{L}s + \frac{1}{{LC}}}}\)

The characteristics equation of series RLC circuit is as follows,

\(s^2 +\frac{R}{L}s+\frac{1}{LC}=0\)

Concept of series RLC Circuit:

A series RLC circuit connected to a voltage source is as shown:

Apply Kirchoff's voltage law to the R-L-C series circuit,

\(Ri + L\frac{{di}}{{dt}} + \frac{1}{C}\mathop \smallint \limits_0^t idt = 0\)

Taking Laplace transform of the above equation,

We get,

\(RI(s)+sLI(s)+\frac{I(s)}{Cs}-i(0^-) +[\frac{I(s)}{s}+\frac{i(0^-)}{s}] =0\)

The initial current i (0^-) is 0. i_c(0^-) is the initial charge across the capacitor and equals - CV₀.

∴ \(RI(s)+sLI(s)+\frac{I(s)}{Cs} +[\frac{I(s)}{Cs}+\frac{V_0}{s}] =0\) 

On simplifying the equation, we get

\(I\left( s \right) = \frac{{{V_0}/L}}{{{s^2} + \frac{R}{L}s + \frac{1}{{LC}}}}\)

The characteristics equation of series RLC circuit is as follows,

\(s^2 +\frac{R}{L}s+\frac{1}{LC}=0\)

From figure;

\({Z_1} = \left( {1 + j1} \right){\rm{\Omega }} = \sqrt 2 \;\angle 45^\circ {\rm{\Omega }}\) 

\({Z_2} = \left( {1 - j1} \right){\rm{\Omega }} = \sqrt 2 \;\angle - 45^\circ {\rm{\Omega }}\) 

The current through Resistor ‘R’ is given as,

\({I_R} = \left( {{{\frac{{{z_1}}}{{{z_1} + z}}}_2}} \right)I\)

\({I_R} = \left[ {\frac{{1 + j1}}{{1 + j1 + 1 - j1}}} \right]\)

\({I_R} = \frac{{\sqrt 2 \angle 45^\circ }}{2} \times 1\angle 0^\circ \) 

\({I_R} = \frac{1}{{\sqrt 2 }}\angle 45^\circ \) 

Voltage across R is given as;

V_R = I_R ⋅ R

\({V_R} = \frac{1}{{\sqrt 2 }}\angle 45^\circ \) 

Concept:

\(kVA = \frac{{kW}}{{cos\phi }}\)

kVAR = (kVA) (sinϕ)

\(sin\phi = \sqrt {1 - {{\cos }^2}\phi } \)

Where,

kVA = Apparent power (S)

kW = Active/True power (P)

kVAR = Reactive power (Q)

cosϕ = Power factor

In an A.C. circuit, true power drawn by the circuit is given by:

True power = Apparent power × Power factor.

Calculation:

Here, input apparent power is 40 kVA and the power factor is 0.8

The power drawn by the circuit is:

(40 × 0.8) kW = 32 kW

Concept of series RLC Circuit:

A series RLC circuit connected to a voltage source is as shown:

Apply Kirchoff's voltage law to the R-L-C series circuit,

\(Ri + L\frac{{di}}{{dt}} + \frac{1}{C}\mathop \smallint \limits_0^t idt = 0\)

Taking Laplace transform of the above equation,

We get,

\(RI(s)+sLI(s)+\frac{I(s)}{Cs}-i(0^-) +[\frac{I(s)}{s}+\frac{i(0^-)}{s}] =0\)

The initial current i (0^-) is 0. i_c(0^-) is the initial charge across the capacitor and equals - CV₀.

∴ \(RI(s)+sLI(s)+\frac{I(s)}{Cs} +[\frac{I(s)}{Cs}+\frac{V_0}{s}] =0\) 

On simplifying the equation, we get

\(I\left( s \right) = \frac{{{V_0}/L}}{{{s^2} + \frac{R}{L}s + \frac{1}{{LC}}}}\)

The characteristics equation of series RLC circuit is as follows,

\(s^2 +\frac{R}{L}s+\frac{1}{LC}=0\)

The capacitive reactance is given by:

\(X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}\)

f = Frequency of the voltage/current applied across the capacitor.

C = Capacitance value

For DC:

A DC signal is a zero frequency signal, i.e. f = 0 Hz

XC for f = 0 will be:

\(X_C=\frac{1}{2\pi (0)C}= \infty \) Ω

Note: A capacitor passes AC signals and blocks DC signals.

For AC:

AC signal is a signal having a particular frequency 'f',

XC for any frequency f is given by:

\(X_C=\frac{1}{2\pi fC}= Finite\)

Observation:

• For DC we conclude that a capacitor provides infinite resistance to the flow of current. Hence no current will flow as current in a capacitive circuit is given by:

          \(I=\frac{V}{X_C}\)

• For AC we observe that the resistance offered by the capacitor is finite. Hence a finite flow of current is possible in a capacitive circuit with AC input.

Correct answer is option (2): an inductor of 1H.

Explanation:

As per the given graph, the current through the element I(t) is lagging, it implies that the element is inductor.

\(I(t)= \frac{V(t)}{Z_{o}}\\ \ I(t)= \frac{sin(t)}{Z_{o}}\)          

The maximum value of the current   \(I(t)_{max} = \frac{1}{\sqrt{2}}\)

\(|Z_{o}|=\frac{|V(t)|}{|I(t)|}\\ =\frac{1}{\frac{1}{\sqrt{2}}}\\ =\sqrt{2}\)

As per the circuit given,

Z_o = R + jωL

\(|Z_{o}|=\sqrt{R^{2}+(j ω L)^{2}}\)

\(\sqrt{2}=\sqrt{1+ ω^{2}L^{2}}\)

Given, ω = 1 rad/sec

\( \sqrt{2} = \sqrt{1+L^{2}}\)

2 = 1 + L²

L = 1 H                           

Concept:

A series RC circuit is drawn as:

The voltage across the capacitor using the voltage division rule is:

\({V_c}\left( ω \right) = V_s\left( {\frac{ \frac{{{1}}}{{jω C}}}{{R + \frac{1}{{jω C}}}}} \right)\)

\({V_c}\left( ω \right) = \frac{{{V_s}}}{{jω C}}\left( {\frac{{jω C}}{{1 + Rjω C}}} \right)\)

\({V_c}\left( ω \right) = \frac{{{V_s}}}{{1 + jω RC}}\) ........ (1)

Where, ω = 2πf

R is the source resistance.

V_s is the source voltage.

Calculation:

Given, f = 1 kHz, R = 50 Ω 

Output peak to peak voltage under open circuit condition is the source peak to peak voltage

Peak to peak V_s = peak to peak V_o = 10 V

Peak V_s = 5 V

When the capacitor is connected across the open circuit terminals, the voltage across the capacitor is given

Peak to peak V_c = 8 V

Peak V_c = 4 V

∴ From equation(1) 

\({1 + jω RC} = \frac{{{V_s}}}{{V_c}}\)

Now considering magnitude on both sides,

\(\sqrt{1 + (ω RC)^2} = \frac{{{V_s}}}{{V_c}}\\ \;\\ω RC=\sqrt{\left(\frac{{{V_s}}}{{V_c}}\right)^2-1}\\\;\\C=\frac{{1}}{{2\pi f R}}\sqrt{\left(\frac{{{V_s}}}{{V_c}}\right)^2-1}\\\;\\C=\frac{{1}}{{2\pi \times 1000\times 50}}\sqrt{\left(\frac{{{5}}}{{4}}\right)^2-1}\)

C = 2.38 μF

The option is randomly marked

Concept:

For series R-C network, If frequency of source increases;

1) Magnitude of current increases

2) Value of capacitive reactance decreases

Calculation: 

\(\left| I \right| = \frac{{\left| V \right|}}{{\sqrt {{R^2} + X_C^2} }}\)

→ If frequency get doubled

1) X_C ↓ & hence |I| ↑

2) V_R ↑ & V_C ↓

Where, |V_R| = |I|. R

|V_C| = |I|.X_C

\(\left\{ {{X_C} = \frac{1}{{{w_C}}}} \right\}\) 

∴ when frequency get doubled;

V_C < V_R

Observation:

1) When frequency increases,

\(\left| Z \right| = \sqrt {{R^2} + X} \)

i.e. impedance value will also get reduces.

2) \(\cos \phi = \frac{R}{Z};\) ∴ power factor will increase.

Concept:

• Since transfer function is given, hence given system is the LTI system.
• For the LTI system, when input is sinusoidal then the output will also be sinusoidal having the same frequency but different amplitude & phase.

Where, \(A' = A \cdot {\left| {H\left( w \right)} \right|_{w = {w_0}}}\)

\(\phi ' = \phi + {\left. {\angle H\left( w \right)} \right|_{w = {w_0}}}\)

Given:

Transfer function; \(H\left( s \right) = \frac{{10\left( {s + 1} \right)}}{{\left( {s + 2} \right)}}\)

Analysis:

Since, \(H\left( s \right) = \frac{{10\left( {s + 1} \right)}}{{\left( {s + 2} \right)}}\) {Replace s = jω}

\(\angle H\left( w \right) = {\tan ^{ - 1}}\left( w \right) - {\tan ^{ - 1}}\left( {\frac{w}{2}} \right)\)

\(\left\{ {\theta = a + jw = {{\tan }^{ - 1}}\left( {\frac{w}{a}} \right)} \right\}\)

As the phase of the system is positive. It can only increase the phase of output i.e. voltage response of the system will lead the current.

Note: System gain only affects the gain of output, not the phase.

From figure;

\({Z_1} = \left( {1 + j1} \right){\rm{\Omega }} = \sqrt 2 \;\angle 45^\circ {\rm{\Omega }}\) 

\({Z_2} = \left( {1 - j1} \right){\rm{\Omega }} = \sqrt 2 \;\angle - 45^\circ {\rm{\Omega }}\) 

The current through Resistor ‘R’ is given as,

\({I_R} = \left( {{{\frac{{{z_1}}}{{{z_1} + z}}}_2}} \right)I\)

\({I_R} = \left[ {\frac{{1 + j1}}{{1 + j1 + 1 - j1}}} \right]\)

\({I_R} = \frac{{\sqrt 2 \angle 45^\circ }}{2} \times 1\angle 0^\circ \) 

\({I_R} = \frac{1}{{\sqrt 2 }}\angle 45^\circ \) 

Voltage across R is given as;

V_R = I_R ⋅ R

\({V_R} = \frac{1}{{\sqrt 2 }}\angle 45^\circ \) 

Concept:

\(kVA = \frac{{kW}}{{cos\phi }}\)

kVAR = (kVA) (sinϕ)

\(sin\phi = \sqrt {1 - {{\cos }^2}\phi } \)

Where,

kVA = Apparent power (S)

kW = Active/True power (P)

kVAR = Reactive power (Q)

cosϕ = Power factor

In an A.C. circuit, true power drawn by the circuit is given by:

True power = Apparent power × Power factor.

Calculation:

Here, input apparent power is 40 kVA and the power factor is 0.8

The power drawn by the circuit is:

(40 × 0.8) kW = 32 kW