Step Response of First Order Circuits MCQ Quiz - Objective Question with Answer for Step Response of First Order Circuits - Download Free PDF

Concept Used:

In an AC circuit with an inductor and capacitor , the impedance of the elements affects the voltage and current distribution.

Impedance (Z): The total opposition offered by circuit components to AC current, given by:

Z = R + jX

The current in an AC circuit follows Ohm's Law for AC circuits :

I = V / Z

Calculation:

Given,

Angular frequency, ω = 80 rad/s

Using the provided values:

Impedance in first branch:

Z₁ = √(125² +50²)= 25√29

Impedance in second branch:

Z₂ = √(80²+ 50²) = 10√89  

⇒ Current in the first branch:

I₁ = (80 / Z₁)

⇒ I₁ = (16 / 5√29) A at 45° leading

⇒ Current in the second branch:

I₂ = 80 / Z₂

⇒ I₂ = (8 / √89) A at 45° lagging

The voltage drop 1H is 

V_H= I₂ × X_L= (8 / √89) × 80 =640/√89 = 67.8 V 

Ans. (2) Sol.

The phase difference (ϕ) between the applied voltage and the current in an A.C. circuit containing a resistor (R) and inductor (with inductive reactance X_L) connected in series is given by: ϕ = tan⁻¹(X_L / R)

Given: Resistance, R = 5 Ω Inductive reactance, X_L = 5 Ω Substitute the values: ϕ = tan⁻¹(5 / 5) ϕ = tan⁻¹(1) ϕ = π / 4 radians

Therefore, the phase difference is π / 4 radians.

• At resonance in a series R-L-C circuit, the inductive reactance (X_L) and capacitive reactance (X_C) are equal and cancel each other out.
• Resonance occurs at the frequency where X_L = X_C.
• Voltage across the capacitor (V_C) can be found using the relationship V_C = I × X_C, where I is the current through the circuit.
• At resonance, the current (I) in the circuit is V_s / R, where V_s is the supply voltage and R is the resistance.
• Reactance of the capacitor X_C = 1 / (ωC), where ω = 1 / √(LC).

To find the resonance frequency (ω):

• ω = 1 / √(LC)
• ω = 1 / √(400 H × 4 F)
• ω = 1 / √(1600)
• ω = 1 / 40 rad/s

At resonance:

• X_L = X_C = 1 / (ωC)

Calculating X_C:

X_C = 1 / (ω × C)
X_C = 1 / (1/40 × 4)
X_C = 40 / 4
X_C = 10 Ω

The current through the circuit (I) is:

• I = V_s / R
• I = 400 V / 5 Ω
• I = 80 A

The voltage across the capacitor (V_C) is:

V_C = I × X_C
V_C = 80 A × 10 Ω
V_C = 800 V

Therefore, the voltage across the capacitor at resonance is 800 V.

Concept:

The impedance of the series RL circuit opposes the flow of alternating current.

The impedance of a series RL Circuit is nothing but the combined effect of resistance (R) and inductive reactance (XL) i.e ωL of the circuit as a whole.

The impedance Zeq in ohms is given by,

Zeq = √(R2+XL2)

Analysis:

If;

Frequency (f) of applied voltage increases → ω ↑  → XL ↑ → Zeq ↑ 

As the value of reactance increases, the phase angle also increases.

Concept:

Admittance is defined as a measure of how easily a circuit or device will allow current to flow through it.

Admittance is the reciprocal (inverse) of impedance, akin to how conductance and resistance are related.

In parallel circuits to calculate equivalent admittance conductance and susceptance are added.

For parallel RL circuit equivalent admittance;

Y = G - jB

where;

Y → Admittance (1/Z)

G→ Conductance(1/R)

B→ Susceptance (1/ωL)

Analysis:

If;

Frequency (f) of applied voltage decreases → ω↓  → X_L↓  → B↑ → Y↓→ Z↑ 

Series RL circuit:

Let the voltage be V_s(t) = Vm sin ωt

The transient current equation is 

\(i_{tr}\left( t \right) = \dfrac{{{V_m}}}{{\sqrt {{R^2}+{ω ^2}{L^2}} }}\ \sin \left( { θ -α} \right) ~{e^{ - \frac{R}{L}t}}\)

where

V_m = Peak value of voltage

θ = Peak value of impedance angle

ω = angular frequency

R = Resistance

L = Inductance

α = instant at which the circuit is closed

θ = Impedance angle

The exponential decay term represents the transient term and the remaining is the steady-state term.

The response is shown below;

In the steady-state, the RLC circuit elements give a response that is in synchronization to the input frequency.

Because transient analysis expressions consist of exponential decay terms.

Size of transient current primarily depends on the instant in the voltage cycle at which circuit is closed i.e. it depends on α value.

Concept:

The current through the inductor in transient is given by:

\(i\left( t \right) = i(\infty ) + \left( {i\left( 0 \right) - i\left( ∞ \right)} \right){e^{ - \frac{{tR}}{L}}}\) ....(1)

i(∞) = Steady-state/final value of the inductor

i(0) = Initial stored current

Also, an inductor does not allow a sudden change in voltage, i.e.

i(0-) = i(0+)

τ = Time Constant

The time constant in the RL series circuit is given by:

\(τ=\dfrac{L}{R}\)

Calculation:

For t < 0:

Initially, for t < 0, the switch is closed for a long time.

The 4 Ω resistor and 30 V source are bypassed and the inductor is short-circuited.

Current flowing in the circuit will be:

\(i_L(0^-)=\dfrac{10}{1}=10~A\)

For t > 0:

After an infinite time when the switch is closed, the inductor is short-circuited and the current will be:

\(i_L(\infty)=\dfrac{10+30}{4+1}=8~A\)

τ = L/R

\(\rm \tau = \frac{{\frac{1}{2}}}{5} = \frac{1}{{10}}\)

From equation (1)

i_L(t) = 8 + [10 - 8] e^-10t

i_L(t) = 8 + 2e^-10t

Given circuit diagram:

The above circuit can be redrawn as, (the above circuit forms a balanced bridge condition)

The equivalent resistance of all the 5 Ω resistors is 5 Ω 

Current in the circuit will be 

i(t) = I e^-4τ

Time constant τ = RC = 5 secs

\(I = \frac{V}{R} = \frac{{10}}{5} = 2A\)

\(i\left( t \right) = 2{e^{ - \frac{t}{5}}} = 2{e^{ - 0.2t}}\)

Energy transferred from the DC source,

\(E = \mathop \smallint \limits_0^\infty 10 \times 2{e^{ - 0.2t}}dt\)

\(= 20\mathop \smallint \limits_0^\infty {e^{ - 0.2t}}\)

\(= 20\left[ {\frac{{{e^{ - 0.2t}}}}{{ - 0.2}}} \right]_0^\infty = 20\left[ {0 + \frac{1}{{0.2}}} \right] = 100\;J\)

Concept:

A capacitor does not allow a sudden change in voltage, i.e.

V_c(0^-) = V_c(0⁺)

Similarly, an inductor does not allow a sudden change in current across it, i.e.

i_L(0^-) = i_L(0⁺)

Analysis:

For t = 0^- the capacitor has no initial voltage, i.e.

V_c(0^-) = 0

∴ V_c(0⁺) = 0V

The steady-state voltage across the capacitor can be obtained by open-circuiting the capacitor as shown:

Using voltage division rule:

\({V_c}\left( \infty \right) = \frac{{5 \times 2}}{{2 + 1}} = \frac{{10}}{3}V\)

The voltage across the capacitor will be:

\({V_C}\left( t \right) = {V_C}\left( \infty \right) - \left( {{V_C}\left( \infty \right) - {V_C}\left( 0 \right)} \right){e^{\frac{{ - t}}{{{R_{eq}}C}}}}\)

\(\therefore {V_C}\left( t \right) = \frac{{10}}{3} - \left( {\frac{{10}}{3} - 0} \right){e^{\frac{{ - t}}{{{R_{eq}}C}}}}\)

\({V_C}\left( t \right) = \frac{{10}}{3}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)\)

τ = R_eq C = Time constant.

R_eq = Equivalent impedance across the capacitor which is obtained as:

\({R_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{{2k}}{3}\)

\(\therefore \tau = \frac{{2k}}{3} \times 1\mu = \frac{2}{3}msec\)

Since the capacitor is connected in parallel to the Resistance R₂, the voltage across R₂ will also be V_C(t).

∴ The required current I will be:

\(I = \frac{{{V_C}\left( t \right)}}{{{R_2}}}\)

\(I = \frac{{10}}{{3 \times 2k}}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)\)

\(I = \frac{5}{3}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right)mA\)

\(\tau = \frac{2}{3}msec\)

Concept:

The Series R-L circuit with DC source is shown below as.

Formula:

The  voltage across the resistor is 

\({V_R} = i.R\)

The voltage across the Inductor is 

\(\begin{array}{l} {V_L} = L.\frac{{di}}{{dt}}\;\\ V = {V_R} + {V_L}\\ V = i.R + L\frac{{di}}{{dt}} \end{array}\)

The formula for alternating current passes through the R-L circuit is

\(i = \frac{V}{R}\left( {1 - {e^{ - \left( {\frac{R}{L}} \right)t}}} \right)\)

Calculation:

\(\begin{array}{l} i = \frac{V}{R}\left( {1 - {e^{ - \left( {\frac{R}{L}} \right)t}}} \right)\\ i = \frac{{120}}{{20}}\left( {1 - {e^{ - \left( {\frac{{20}}{8}} \right)0.6}}} \right)\\ i = 6\left( {1 - {e^{ - 1.5}}} \right)\\ i = 4.66\;A \end{array}\)

• At resonance in a series R-L-C circuit, the inductive reactance (X_L) and capacitive reactance (X_C) are equal and cancel each other out.
• Resonance occurs at the frequency where X_L = X_C.
• Voltage across the capacitor (V_C) can be found using the relationship V_C = I × X_C, where I is the current through the circuit.
• At resonance, the current (I) in the circuit is V_s / R, where V_s is the supply voltage and R is the resistance.
• Reactance of the capacitor X_C = 1 / (ωC), where ω = 1 / √(LC).

To find the resonance frequency (ω):

• ω = 1 / √(LC)
• ω = 1 / √(400 H × 4 F)
• ω = 1 / √(1600)
• ω = 1 / 40 rad/s

At resonance:

• X_L = X_C = 1 / (ωC)

Calculating X_C:

X_C = 1 / (ω × C)
X_C = 1 / (1/40 × 4)
X_C = 40 / 4
X_C = 10 Ω

The current through the circuit (I) is:

• I = V_s / R
• I = 400 V / 5 Ω
• I = 80 A

The voltage across the capacitor (V_C) is:

V_C = I × X_C
V_C = 80 A × 10 Ω
V_C = 800 V

Therefore, the voltage across the capacitor at resonance is 800 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

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

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

Concept:-  A coil is made up of inductive reactance and Resistance.

i.e. When it is connected to the DC supply, the inductor will act as a short circuit

Hence, Resistance of the coil =  Ω

For AC supply, the inductor will offer finite reactance that will depend on the frequency of the supply.

i.e, X_L = ωL

Now,

Z = V/I = 32/5 = 6.4 Ω

We know that,

Power Factor = R/Z = 5/6.4 = 0.78

And,

X² = Z² - R²

Hence,

X = \(\sqrt{6.4^2-5^2}\) = 4 Ω

Concept:

For a series RL circuit, the net impedance is given by:

Z = R + j (XL)

XL = Inductive Reactance given by:

XL = ωL

The magnitude of the impedance is:

\(|Z|=\sqrt{R^2+X_L^2}\)

Calculation:

With R = 3 Ω and XL = 4 Ω, we get:

\(|Z|=\sqrt{(3)^2+(4)^2}\)

\(|Z|=\sqrt{25}\)

|Z| = 5 Ω

Important Notes:

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

\(X_C=\frac{1}{\omega C}\)

The magnitude of the impedance will be:

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

Concept:

Admittance is defined as a measure of how easily a circuit or device will allow current to flow through it.

Admittance is the reciprocal (inverse) of impedance, akin to how conductance and resistance are related.

In parallel circuits to calculate equivalent admittance conductance and susceptance are added.

For parallel RL circuit equivalent admittance;

Y = G - jB

where;

Y → Admittance (1/Z)

G→ Conductance(1/R)

B→ Susceptance (1/ωL)

Analysis:

If;

Frequency (f) of applied voltage decreases → ω↓  → X_L↓  → B↑ → Y↓→ Z↑