Source Free Circuits MCQ Quiz - Objective Question with Answer for Source Free Circuits - Download Free PDF

Concept:

Phase Difference in an R-L Circuit: In a series AC circuit containing a resistor (R) and an inductor (L), the phase difference (ϕ) between the current (I) and voltage (V) is given by:

• Formula: tanϕ = X_L / R
• Here,
  • X_L = Inductive reactance = ωL = 2πfL
  • ϕ = Phase angle (degrees)
  • R = Resistance (Ω)
  • L = Inductance (H)
  • f = Frequency (Hz)

Calculation:

Given,

Inductance, L = 25.48 mH = 25.48 × 10^-3 H

Resistance, R = 8 Ω

Frequency, f = 50 Hz

⇒ Inductive reactance, X_L = 2πfL

⇒ X_L = 2 × 3.14 × 50 × (25.48 × 10^-3)

⇒ X_L = 2 × 3.14 × 50 × 0.02548

⇒ X_L ≈ 8 Ω

⇒ Phase difference, tanϕ = X_L / R

⇒ tanϕ = 8 / 8 = 1

⇒ ϕ = tan^-1(1) = 45°

∴ The phase difference between current and voltage in this circuit is 45°.

Answer : 2

Solution :

Power factor = \(\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{2}}}\)

Given : X_L = R

∴ P₁ = \(\frac{\mathrm{R}}{\sqrt{2 \mathrm{R}^{2}}}=\frac{1}{\sqrt{2}}\)

When C is connected, the circuit becomes series LCR circuit. 

∴ Power factor = \(\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}}}\)

Given : X_C = X_L

∴ P₂ = \(\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}}}\) = 1

∴ P1 ∶ P2 = 1 ∶ √2

RC network are smaller in size, have lower losses and very cheaper.

EXPLANATION:

Concept Used:

The charge in an oscillating circuit decays exponentially according to the equation:

Q = Q₀ e^-Rt/2L

where:

Q₀ = Initial charge

R = Resistance = 1.5 Ω

L = Inductance = 12 mH = 12 × 10^-3 H

ln(2) = 0.693

We need to determine the time t when Q = 0.5 Q₀.

Calculation:

Substituting the given condition Q = 0.5 Q₀:

0.5 Q₀ = Q₀ e^-Rt/2L

Dividing by Q₀:

0.5 = e^-Rt/2L

Taking natural logarithm on both sides:

ln(0.5) = -Rt / 2L

ln(1/2) = -Rt / 2L

ln(2) = Rt / 2L

Rearranging for t :

t = (2L ln 2) / R

Substituting values:

t = (2 × 12 × 10^-3 × 0.693) / 1.5

t = 11.09 × 10^-3 s

t = 11.09 ms

∴ The correct answer is 11.09 ms.

Concept

The impedance of a series RL circuit is given by:

\(Z=R+jX_L\)

\(Z=R+jω L\)

The magnitude of the impedance is:

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

Calculation

Given, R = 10 Ω and L = 0.056 H

f = 50 Hz

ω = 2πf = 314 radian

X_L = ω × L

XL = 314 × 0.056 = 17.584 Ω 

\(\left | Z \right |=\sqrt{(10)^2+(17.584)^2}\)

\(\left | Z \right |=20.23\space \Omega\)

The correct answer is option

Concept:

The Bandwidth of Series RLC CIrcuit is given as

B.W = \(R \over L\)  rad/s

Where R is the resistance in ohm

L is the Inductance in henry

Calculation:

Given

R = 1000 Ω, L = 100 mH

B.W = \(1000 \over 100 × 10^{-3}\)

= 10 × 10³  rad/s

Mistake Points

• Bandwidth= \(\frac{R}{L} \ rad/s\)
• Bandwidth=  \(\frac{R}{2\pi L}\ Hz\)

Concept:

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

R = Resistance

XL = Inductive reactance

XC ­= Capacitive reactance

Net reactance of the circuit is given by

X = |XL – XC|

Calculation:

Given that,

R = 20 Ohms, XL = 30 Ohms and XC = 10 Ohms

The net reactance of the circuit will be:

X = |30 - 10|

X = 20 Ω

The correct answer is option 3):( Purely resistive circuit)

Concept:

In a series RLC circuit at resonance, inductive reactance (XL) is equal to capacitive reactance (XC). The circuit becomes purely resistive.

The capacitive reactance (XC) is given by

X_C = \(1 \over 2π f C\)

C = ∞ 

XC = 0

when 

The inductive reactance ( X_L)  is given 

X_L = 2 π f L

 L = 0.

XL = 0

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

where 

Z is the impedance of the circuit

R is the resistance of the circuit

f is the frequency

The correct answer is option 3):(0 to 1 )

Concept:

• In a Series R-C circuit, for maximum power Rc = Xc

Power factor = cos⁡ϕ = R/|Z|

|Z| = \( \sqrt {R_2+X_{c2}}\)

= \(\sqrt{R_2+R_2 } \) =  \(\sqrt 2\)R

Power factor cos⁡ϕ = \( R\over \sqrt{2}R\) =  \(1 \over \sqrt{2} \)

• The power factor corresponding to maximum power is 0.707 lead.
• The power factor is leading because the circuit has capacitive reactance.
• The range of operating power factor for any given RC series circuit is 0 to 1

Capacitive reactance:

The Capacitive reactance of a capacitor is given by:

\({X_C} = \frac{1}{{ω C}}\)

With ω = 2πf

\({X_C} = \frac{1}{{2π fC}}\)

Calculation:

For a DC supply of 10 V, the frequency is zero, i.e. f = 0 Hz.

For a DC supply of 10 V, under steady-state, i.e. at t → ∞, the capacitor will act as an open circuit because it will offer infinite resistance.

∴ The steady-state current will be 0 A.

At t < 0, the circuit is as shown below.

The inductor acts as short circuit and the current flows through the inductor is

\({i_L}\left( {{0^ - }} \right) = \frac{{10}}{4} = 2.5\;A\)

At t > 0, the circuit becomes as shown below:

i_L(0⁺) = i_L(0^-) = 2.5 A

At t = ∞, as there is no source, the current will be zero, i.e.

i_L (∞) = 0

Time constant (τ) = L/R = 5/10 = 0.5

The transient equation for the current flows through the inductor is,

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

\( = 0 + \left( {2.5 - 0} \right){e^{ - 2t}} = 2.5{e^{ - 2t}}\)

The voltage across the inductor is, \({V_L}\left( t \right) = L\frac{{d{i_L}}}{{dt}} = - 25{e^{ - 2t}}\)

The correct answer is (option 3) i.e. 0.632

Concept:

Time constant is defined as the time in which the current reaches 63.2% of its steady-state value.

Time constant of R-C circuit is,  τ = RC  sec. 

Where, 

τ = Time constant

R= Resistance of the circuit,  C = Capacitance of the circuit

The above curve is the charging state of the capacitor in a one-time constant

At one time constant Capacitor is charged 63% of its full capacity.

The above curve is discharging state of the capacitor in a one-time constant.

At one time-constant capacitor loses 63% of its initial full charged voltage. So it contains only 37% of its full capacity voltage.

Since voltage V is related to the charge on a capacitor given by the equation, V_c = \(\frac{Q}{C}\), the voltage across the capacitor (V_c) at any instant in time during the charging period is given as:\(V_c = V_s(1-e^{\frac{-t}{RC}})\)

Where:

• V_c is the voltage across the capacitor
• V_s is the supply voltage
• It is the elapsed time since the application of the supply voltage
• RC is the time constant of the RC charging circuit

Additional InformationTime constant of R-L circuit is, L/R

Concept:

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:

XC = 1/ωC

Impedance = Resistance + j Reactance

The magnitude of the impedance is given by:

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

Application:

Given:

R = 50 Ω,

XC = 90 Ω and,

XL = 30 Ω

From the above concept, impedance (Z) is given as,

Z = R + j (XL - XC)

or, Z = 50 + j(30- 90) = (50 - j60) Ω

The correct answer is: Bandwidth is the ratio of quality factor to the resonant frequency, as this statement is not true with regard to parallel RLC circuit.

Concept:

• The bandwidth of the RLC circuit is defined as the range of frequencies for which circuit output voltage (or) current value equals 70.7 % of its maximum amplitude, which will occur at the resonant frequency.
• Bandwidth is defined as the size of the frequency range that is passed or rejected by the tuned circuit.
• Bandwidth can be given in terms of resonant frequency and quality factor and the formula for calculating bandwidth is the same for series and parallel circuits.

The characteristic equation is given as,

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

The formula for calculating bandwidth is the same for series and parallel circuits.

The bandwidth of a parallel RLC network is given as

\({\rm{BW}} = \frac{{{{\rm{\omega}}_{\rm{r}}}}}{{\rm{Q}}}=\frac{1}{RC}\)\(=\frac{R}{L}\)

• The bandwidth can be given by the ratio of resonant frequency to quality factor.
• Bandwidth is independent of resonant frequency term, though it depends upon circuit parameters (R,L,C).
• At resonance, the imaginary part of the net impedance becomes zero, which makes the input impedance a real quantity. Also, at resonance, the input impedance for a parallel resonant circuit attains a maximum value.
• This is opposite to a series resonance where at resonance, the impedance attains a minimum value.

In a series RLC circuit, the impedance is given by

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

At resonance, the magnitude of inductive reactance is equal to the magnitude of capacitive reactance.

The magnitude of XL = XC

At this condition, Z = R.

Hence at resonance, the impedance is purely resistive and it is minimum.

Current in the circuit, I = V/Z

As impedance is minimum the current is maximum.

As impedance is purely resistive, the power factor is unity.

Important Points:

• In a purely inductive circuit, the current lags the voltage by 90° and the power factor is zero lagging
• In a purely capacitive circuit, the current leads the voltage by 90° and the power factor is zero leading

The plot of the frequency response of the series circuit is as shown in the figure: