Variation of Voltages Across R/L/C with Frequency MCQ Quiz - Objective Question with Answer for Variation of Voltages Across R/L/C with Frequency - Download Free PDF

As explained in the previous response, the correct answer is 2) 0.

Explanation

• At resonance in a series RLC circuit, the inductive reactance (X_L) and capacitive reactance (X_C) are equal in magnitude but opposite in phase
• This results in the voltages across the inductor (V_L) and the capacitor (V_C) being equal in magnitude and 180 degrees out of phase.
• Therefore, the combined voltage across the inductor and capacitor is zero.

Concept

The source voltage for a series RLC circuit is given by:

\(V_S=V_R+j(V_L-V_c)\)

Calculation

Given that, V_R = V_L = V_C = 30 V

\(V_S=30+j(30-30)\)

V_S = 30 V

Concept:

Impedance and Phase Angle in an LCR Circuit:

• In a series LCR circuit with an alternating voltage source, the total impedance Z is given by Z = √(R² + (XL - XC)²), where R is the resistance, XL is the inductive reactance (XL = 2πfL), and XC is the capacitive reactance (XC = 1 / 2πfC).
• The phase angle φ between the voltage and the current is given by tan φ = (XL - XC) / R.
• When the voltage leads the current by 45°, tan 45° = 1, so (XL - XC) / R = 1.
• By substituting XL = 2πfL and XC = 1 / 2πfC into the equation, we can solve for the value of L.

Calculation:

Given, the phase angle φ = 45° and tan 45° = 1.

(XL - XC) / R = 1

Substituting XL = 2πfL and XC = 1 / 2πfC,

[(2πfL) - (1 / 2πfC)] / R = 1

Multiplying both sides by R,

2πfL - 1 / 2πfC = R

Rearranging for L,

L = (R + 1 / 2πfC) / 2πf

Thus, the value of L is (1 + 2πfCR) / 4π²f²C.

∴ The value of L is (1 + 2πfCR) / 4π²f²C. Hence, option 1) is correct.

Concept:

For a series RLC circuit, the resonant frequency is given as:

\({\omega _0} = \frac{1}{{\sqrt {LC} }}\)

Application:

Given L = 50 mH and C = 5 μF

\({\omega _0} = \frac{1}{{\sqrt {50m \times 5\mu } }} = 2k\;rad/sec\)

 At resonance, the net impedance will be purely resistive.

∴ The current flowing in the circuit at resonance will be:

\(I = \frac{V}{R}\)

\(I = \frac{{60\angle 0}}{{10k}} = 6\angle 0^\circ \;mA\)

Now, the voltage across the capacitor at resonance will be:

\({V_c} = I\;\left( {\frac{1}{{j{\omega _0}C}}} \right) = \frac{I}{{{\omega _0}C\angle 90^\circ }}\)

Putting on the respective values, we get:

\({V_c} = \frac{{6m\;\angle 0}}{{2k \times 5\mu \;\angle 90^\circ }} = \frac{6}{{10}}\;\angle - 90^\circ \)

V_c = 0.6 ∠-90° 

Note:

The voltage across the inductor and the capacitor will be finite and of equal magnitude. Please note that the voltage across the LC combination is zero, as LC becomes a short circuit at resonance. 

Concept:

For a series RLC circuit, as shown, the net magnitude/amplitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_C)^2}\)

V_R = Voltage across the resistor

V_L = Voltage across the inductor 

V_C = Voltage across the capacitor

Analysis:

At resonance, the net impedance is purely resistive resulting in V_L = V_C 

This is explained as shown:

Given V_L = V_C = 50 V

V_R = 120 V

The net amplitude/magnitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_L)^2}\)

\(V_{net}=\sqrt{V_R^2}=V_R=120~V\)

Concept:

For a series RLC circuit, as shown, the net magnitude/amplitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_C)^2}\)

V_R = Voltage across the resistor

V_L = Voltage across the inductor 

V_C = Voltage across the capacitor

Analysis:

At resonance, the net impedance is purely resistive resulting in V_L = V_C 

This is explained as shown:

Given V_L = V_C = 50 V

V_R = 120 V

The net amplitude/magnitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_L)^2}\)

\(V_{net}=\sqrt{V_R^2}=V_R=120~V\)

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

Z = R + j (X_L - X_C)

X_L = Inductive Reactance given by:

X_L = ωL

X_C = Capacitive Reactance given by:

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

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

If X_L > X_C, then the nature of the circuit is inductive.

If  XC > XL, then the nature of the circuit is capacitive.

Concept:

• Superposition theorem applies to AC circuits in the same way they apply to the DC circuit.
• If a linear circuit is excited by several sinusoidal sources, all having the same frequency ω, then superposition may be used to evaluate the circuit current/voltage with several sources.
• But, if a circuit has two or more sources acting at different frequencies, then superposition must be used.

​

Application:

Given: v_i(t) = 2cos (200t) + 4 sin (500t)

Case 1: when input is 2 cos 200t

At ω = 200 rad/sec, the series combination of acts like short circuit because L and C combination is at resonance.

∴ \({{\rm{v}}_0}{\rm{}} = {\rm{}}{{\rm{v}}_{{\rm{in}}}}{\rm{}} = {\rm{}}2{\rm{\cos{200{\rm{t}}}}}\)

Case 2: When the input is 4 sin500t

At ω = 500 rad/sec, the parallel combination of L and C are acts like an open circuit because of resonance. So, there is no flow of current.

So again \({{\rm{v}}_0}{\rm{}} = {\rm{}}{{\rm{v}}_{{\rm{i}}}}{\rm{}} = {\rm{}}4{\rm{sin\;}}500{\rm{t}}\)

∴ The overall output will be:

\({{\rm{v}}_0}{\rm{\;}} = {\rm{\;}}2{\rm{cos\;}}200{\rm{t\;}} + {\rm{\;}}4{\rm{sin}}500{\rm{t}}\)

Given circuit is in resonance,

\(\begin{array}{l} \Rightarrow {f_c} = \frac{1}{{2\pi \sqrt {LC} }}\;or\;{\omega _c} = \frac{1}{{\sqrt {LC} }}\\ {\omega _c} = \frac{1}{{\sqrt {0.1 \times 0.1 \times {{10}^{ - 6}}} }} \end{array}\)

= 10000 rad/s

⇒ X_L = X_C = j 0.1 × 10000

= j 1000Ω

The current through the circuit \(i = \frac{{1V}}{{10{\rm{\Omega }}}}\)

= 0.1 A

∴ V_C = |X_C . i|

= |0.1 × j 10³|

= 100 V

Concept:

In a series RLC circuit; if the voltage and current are in phase, then this indicates that the inductive and capacitive reactance are both equal and the circuit is in resonance.

Given, I̅ and V̅ are in phase.

i.e. X_C = X_L.

Calculation:

At resonance,

\({\omega _o} = \frac{1}{{\sqrt {LC} }}\)    ---(1)

The voltage across the capacitor is:

\(|V_c|=|X_c|\times|I|\)    ---(2)

Similarly, the voltage across the resistor is:

 \( \left| {{V_R}} \right| =|I|\times{R}\)    ---(3)

Using Equations 1,2 and 3 we get:

\(\frac{{\left| {{V_C}} \right|}}{{\left| {{V_R}} \right|}} = \frac{{{X_C}}}{R} \)

\(\Rightarrow\frac{1}{{{\omega _o}C \times R}} = \frac{{\sqrt {LC} }}{{C \times R}} = \sqrt {\frac{L}{C}} .\frac{1}{R}\)

Given,

R = 5Ω

L = 5 H

and C = 5 F

Concept

The source voltage for a series RLC circuit is given by:

\(V_S=V_R+j(V_L-V_c)\)

Calculation

Given that, V_R = V_L = V_C = 30 V

\(V_S=30+j(30-30)\)

V_S = 30 V

As explained in the previous response, the correct answer is 2) 0.

Explanation

• At resonance in a series RLC circuit, the inductive reactance (X_L) and capacitive reactance (X_C) are equal in magnitude but opposite in phase
• This results in the voltages across the inductor (V_L) and the capacitor (V_C) being equal in magnitude and 180 degrees out of phase.
• Therefore, the combined voltage across the inductor and capacitor is zero.

Concept:

For a series RLC circuit, as shown, the net magnitude/amplitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_C)^2}\)

V_R = Voltage across the resistor

V_L = Voltage across the inductor 

V_C = Voltage across the capacitor

Analysis:

At resonance, the net impedance is purely resistive resulting in V_L = V_C 

This is explained as shown:

Given V_L = V_C = 50 V

V_R = 120 V

The net amplitude/magnitude of the source voltage is given by:

\(V_{net}=\sqrt{V_R^2+(V_L-V_L)^2}\)

\(V_{net}=\sqrt{V_R^2}=V_R=120~V\)

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

Z = R + j (X_L - X_C)

X_L = Inductive Reactance given by:

X_L = ωL

X_C = Capacitive Reactance given by:

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

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

If X_L > X_C, then the nature of the circuit is inductive.

If  XC > XL, then the nature of the circuit is capacitive.

Concept:

• Superposition theorem applies to AC circuits in the same way they apply to the DC circuit.
• If a linear circuit is excited by several sinusoidal sources, all having the same frequency ω, then superposition may be used to evaluate the circuit current/voltage with several sources.
• But, if a circuit has two or more sources acting at different frequencies, then superposition must be used.

​

Application:

Given: v_i(t) = 2cos (200t) + 4 sin (500t)

Case 1: when input is 2 cos 200t

At ω = 200 rad/sec, the series combination of acts like short circuit because L and C combination is at resonance.

∴ \({{\rm{v}}_0}{\rm{}} = {\rm{}}{{\rm{v}}_{{\rm{in}}}}{\rm{}} = {\rm{}}2{\rm{\cos{200{\rm{t}}}}}\)

Case 2: When the input is 4 sin500t

At ω = 500 rad/sec, the parallel combination of L and C are acts like an open circuit because of resonance. So, there is no flow of current.

So again \({{\rm{v}}_0}{\rm{}} = {\rm{}}{{\rm{v}}_{{\rm{i}}}}{\rm{}} = {\rm{}}4{\rm{sin\;}}500{\rm{t}}\)

∴ The overall output will be:

\({{\rm{v}}_0}{\rm{\;}} = {\rm{\;}}2{\rm{cos\;}}200{\rm{t\;}} + {\rm{\;}}4{\rm{sin}}500{\rm{t}}\)