AC Voltage Applied to a Series LCR Circuit MCQ Quiz - Objective Question with Answer for AC Voltage Applied to a Series LCR Circuit - Download Free PDF

Last updated on Jun 26, 2025

Latest AC Voltage Applied to a Series LCR Circuit MCQ Objective Questions

AC Voltage Applied to a Series LCR Circuit Question 1:

An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω. is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ______ mH.

(take π2 = 10) 

Answer (Detailed Solution Below) 100

AC Voltage Applied to a Series LCR Circuit Question 1 Detailed Solution

Calculation:

The amplitude will be maximum when there will be resonance condition

The value of resonance frequency is f = 1 / (2π√(L C))

⇒ 2000 Hz = 1 / (2π√(L × 62.5 × 10–9))

⇒ L = 1 / (4π2 × 20002 × 62.5 × 10–9) = 0.1 H = 100 mH

AC Voltage Applied to a Series LCR Circuit Question 2:

An ideal inductor 𝐿 is connected in series to a 150Ω resistor as shown in the circuit (inset). When the circuit is driven by a battery 𝐵1, the voltage across the resistor as a function of time, as measured by an oscilloscope, is shown in the plot.
qImage682c5e616d65ab6652733c3f
Based on this observation, the estimated value of 𝐿 is closest to 

  1. 50 mH 
  2. 300 mH 
  3. 450 mH 
  4. 150 mH 

Answer (Detailed Solution Below)

Option 4 : 150 mH 

AC Voltage Applied to a Series LCR Circuit Question 2 Detailed Solution

Solution:  

−V + L(dI/dt) + IR = 0     (V is source voltage of B1)

dI/dt + (R/L) · I = V/L

Integrating factor  = eRt/L,

⇒ I · eRt/L = ∫ (V/L) · eRt/L dt + c

⇒ I = V/(LR) + c · e−Rt/L

⇒ V = V + c · e−Rt/L

At t = 0, V = 0 → 0 = V + c → c = −V

V = 25 · [1 − e−Rt/L]     (From graph: At t → ∞, V = 25)

At t = 1, V = 16:

16 = 25 · [1 − e−R/L]

⇒ L = R

AC Voltage Applied to a Series LCR Circuit Question 3:

To an AC power supply of 220 V at 50 Hz, a resistor of 20 Ω, a capacitor of reactance 25 Ω and an inductor of reactance 45 Ω are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:

  1. 7.8 A and 30°
  2. 7.8 A and 45°
  3. 15.6 A and 30°
  4. 15.6 A and 45°

Answer (Detailed Solution Below)

Option 2 : 7.8 A and 45°

AC Voltage Applied to a Series LCR Circuit Question 3 Detailed Solution

Correct option is: (2) 7.8 A and 45°

XL = 45 W, XC = 25 W, R = 20 W

⇒ I = 220 / √((XL − XC)2 + R2) = 220 / √((45 − 25)2 + 202)

⇒ = 220 / (2√2) = 11 / √2 = 7.779 A

⇒ tan φ = (XL − XC) / R = (45 − 25) / 20 = 1

⇒ φ = 45°

AC Voltage Applied to a Series LCR Circuit Question 4:

These are connected to variable DC voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three circuits) in different ways as shown in Column II. When a current (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage V1 and V2 (indicated in circuits) are related as shown in Column I.

Column I Column II
(A) I ≠ 0, V1 ∝ I (p)
qImage681b099ec9c0d259d407e7a7
(B) I ≠ 0, V2 > V1 (q)
qImage681b099ec9c0d259d407e80f
(C) V1 = 0, V2 = V (r)
qImage681b099fc9c0d259d407e812
(D) I ≠ 0, V2 ∝ I (s)
qImage681b099fc9c0d259d407e813
 

  1.  A → (q, r, s), B → ( r, s), C → (p, q), D → (q, r, s )
  2.  A → (r, s), B → (q, r, s), C → (p, q), D → (q, r, s )
  3.  A → (r, s), B → (q, r, s), C → (p, q, s), D → (q, r)
  4.  A → (q, r, s), B → ( r, s), C → (p, q), D → (p, r, s )

Answer (Detailed Solution Below)

Option 2 :  A → (r, s), B → (q, r, s), C → (p, q), D → (q, r, s )

AC Voltage Applied to a Series LCR Circuit Question 4 Detailed Solution

Calculation:

In circuit (p): Under steady state, the capacitor will act as an infinite impedance and the inductor will act as a zero impedance. Thus:

I = 0

V1 = 0

V2 = V

In circuit (q): The inductor will act as zero resistance in steady state, giving us:

I = V / R = V / 2

V1 = 0

V2 = V

In circuit (r): The inductive reactance XL and total impedance Z are:

XL = ωL = 2πνL = 1.88 Ω

Z = √(XL2 + R2) = 2.75 Ω

Thus,

I = V / Z ≠ 0

V1 = XLI = 1.88I

V2 = RI = 2I

V2 > V1

In circuit (s): Inductive reactance XL, capacitive reactance XC, and impedance Z are:

XL = 1.88 Ω

XC = 1 / (ωC) = 1061 Ω

Z = XC − XL = 1059 Ω

So the current is:

I = V / Z ≠ 0

V1 = XLI = 1.88I

V2 = XCI = 1061I

V2 > V1

Answer:  A → (r, s), B → (q, r, s), C → (p, q), D → (q, r, s )

AC Voltage Applied to a Series LCR Circuit Question 5:

In a series LCR circuit R=300Ω,L=0.9H,C=2μF,ω=1000 rad/s. The impedance of the circuit is?

  1. 500Ω
  2. 1300Ω
  3. 400Ω
  4. 900Ω
  5. 1100 Ω 

Answer (Detailed Solution Below)

Option 1 : 500Ω

AC Voltage Applied to a Series LCR Circuit Question 5 Detailed Solution

Calculation:

Z=R2+(ωL1ωC)

=(300)2+(1000×0.911000×2×106)2

Z=(300)2+(400)2500Ω

The correct option is (a)

Top AC Voltage Applied to a Series LCR Circuit MCQ Objective Questions

The resonant frequency of a RLC circuit is equal to __________.

  1. 1 / (LC)
  2. 1 / (LC)2
  3. √(LC)
  4. 1 / √(LC)

Answer (Detailed Solution Below)

Option 4 : 1 / √(LC)

AC Voltage Applied to a Series LCR Circuit Question 6 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and the inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

CALCULATION:

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2

  • Inductive reactance,

⇒ XL = Lω  

  • Capacitive reactance

Xc=1Cω

  • Resonance will take place when XL = XC.

⇒ XL = XC

Lω=1Cω

ω=1LC

At resonance the impedance is:

  1. purely resistive
  2. purely capacitative
  3. purely inductive
  4. None of these

Answer (Detailed Solution Below)

Option 1 : purely resistive

AC Voltage Applied to a Series LCR Circuit Question 7 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

EXPLANATION :

  • Resonance of an LCR circuit is the frequency at which capacitive reactance is equal to the inductive reactance, Xc = XL, then the value of impedance is given by

Z=R2+(XLXC)2

Z=R2+0

Z=R

  • At resonance, the value of impedance is purely resistive. Hence, option 1 is the answer

The mathematical form of the resonant frequency of a LCR circuit is equal to 

  1. 12π(LC)
  2. 12π(LC)2
  3. 2π(LC)
  4. 12πLC

Answer (Detailed Solution Below)

Option 4 : 12πLC

AC Voltage Applied to a Series LCR Circuit Question 8 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

CALCULATION:

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2

  • Inductive reactance,

⇒ XL = Lω  

  • Capacitive reactance

Xc=1Cω

  • Resonance will take place when XL = XC.

⇒ XL = XC

Lω=1Cω

ω=1LC

As we know, ω = 2πf

Where f = frequency

f=12πLC

A 220V, 50 Hz ac source is connected in series to a 30 Ω resistor, an inductor, and a capacitor, each having 200 Ω inductive reactance and 160 Ω capacitive reactance, respectively. The voltage drop across the resistor is ______. 

  1. 132 V
  2. 52 V
  3. 22 V
  4. 92 V

Answer (Detailed Solution Below)

Option 1 : 132 V

AC Voltage Applied to a Series LCR Circuit Question 9 Detailed Solution

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Given:

R=30Ω,XL=200Ω,XC=160Ω,V=220V

Concept:

Impedance is the opposition to the alternating current presented by the combined effect of resistance and reactance in a circuit.

Impedance is given by the formula,

  • Z=R2+(XLXC)2

 

Where R is Resistance,

 XL is inductive reactance,

XC is capacitive reactance

Explanation:

  • Z=R2+(XLXC)2
  • Z=302+(200160)2=302+402=2500=50Ω
  • Z=302+402=2500=50Ω

 

Current, I=VZ=22050=225A

Potential drop across the resistor is Vd=IR=225×30=132V

The correct answer is Option-1-132V.

In a series RLC circuit, the values of R, L, and C are 1000 Ω, 4 H, and 10-6 F respectively. What will happen to the resonant frequency of the circuit if the value of R is decreased by 20 Ω ?

  1. It will decrease
  2. It will increase 
  3. Initially it will increase and then decrease
  4. No change

Answer (Detailed Solution Below)

Option 4 : No change

AC Voltage Applied to a Series LCR Circuit Question 10 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

  • When the LCR circuit is set to resonance, the resonant frequency is

f=12π1LC

CALCULATION:

  • From the above, it is clear that the resonant frequency is given by

f=12πLC

  • From the above equation, it is clear that the resonant frequency is independent of resistance, so if the value of R is decreased by 20 Ω  then there will be no change in the value of resonant frequency. Therefore the correct answer is option 4.   

For L-C-R series A.C. circuit, in resonating conditions which is true?

  1. Minimum current
  2. Minimum impendance
  3. Power loss minimum
  4. Minimum power factor

Answer (Detailed Solution Below)

Option 2 : Minimum impendance

AC Voltage Applied to a Series LCR Circuit Question 11 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and the inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

  • The resonant frequency of a series LCR circuit is given by

ν=12πLC

Reactance: 

  • It is basically the inertia against the motion of the electrons in an electrical circuit.
  • There are two types of reactance:

    1. Capacitive reactance (XC) (Ohms is the unit)

    2. Inductive reactance (XL) (Ohms is the unit)

CALCULATION:

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2

  • For resonating condition,

XL=XC

Z=R

  • Hence the impedance is minimum for the resonating condition.
  • Hence, option 2 is correct.

For a series LCR circuit at resonance, the statement which is not true is

  1. Peak energy stored by a capacitor = peak energy stored by an inductor
  2. Average  power = apparent power
  3. Wattles current is zero
  4. Power factor is zero

Answer (Detailed Solution Below)

Option 4 : Power factor is zero

AC Voltage Applied to a Series LCR Circuit Question 12 Detailed Solution

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CONCEPT:

LCR CIRCUIT:

F1 P.Y 7.5.20 Pallavi D2

  •  An LCR Circuit is an electrical circuit consisting of Inductor (L), Capacitor (C)Resistor (R) it can be connected either parallel or series.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =inductive reactance and XC = capacitive reactive

  • At resonance Xc = XL,  the frequency of oscillation is given by

ν=12πLC

Where L = Inductance, C = Capacitance

EXPLANATION :

  • The power factor of LCR is given by

⇒ P  = VI cosϕ  ----(1)

Cosϕ=RZ

at resonance R = Z

Cosϕ=ZZ=1

Substituting the above  value in equation 1

⇒ P = VI

  • From the above equation, it is clear that the power factor is not zero. So, option 4 is wrong
  • Hence, option 4 is the answer.

Power factor in an AC circuit is given by: 

  1. Z/R
  2. R/Z
  3. RXL
  4. RXC

Answer (Detailed Solution Below)

Option 2 : R/Z

AC Voltage Applied to a Series LCR Circuit Question 13 Detailed Solution

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CONCEPT:

  • Power Factor: The power factor of series-connected L and R circuit in AC voltage is

cos ϕ = R/Z

where R is resistance and Z is overall impedance.​

EXPLANATION:

In an AC circuit, the power factor is given by:

cos ϕ = R/Z

So the correct answer is option 2.

Additional Information

  • Voltage in AC: In an AC voltage source, the voltage of the source keeps changing with time and is defined as

V = V0 sin ωt

where V is the voltage at any time t, V0 is the max value of voltage, and ω is the angular frequency.

  • With AC source capacitance, In a series combination of a resistor (R), an inductance (L)

Inductance Reactance is defined as:

 XL=ωL=2πfL

Where ω is the angular frequency, ƒ is the frequency in Hertz, C is the AC capacitance in Farads, and XL is the Inductive Reactance in Ohms,.

F1 J.K 8.9.20 Pallavi D1

Overall Impedance (Z) in a series combination of a resistor (R), an inductance (L):

Z=R2+ωL2

where R is resistance and ωL is Inductance reactance.

The phase difference between the current and voltage in L-C-R circuit at resonance is:

  1. 0
  2. π2
  3. π

Answer (Detailed Solution Below)

Option 1 : 0

AC Voltage Applied to a Series LCR Circuit Question 14 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and the inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

  • The resonant frequency of a series LCR circuit is given by

ν=12πLC

If ϕ is the phase difference between the current and voltage in L-C-R circuit, then,

tanϕ=XLXCR=VLVCVR

​Reactance: 

  • It is basically the inertia against the motion of the electrons in an electrical circuit.
  • There are two types of reactance:

    1. Capacitive reactance (XC) (Ohms is the unit)

    2. Inductive reactance (XL) (Ohms is the unit)​

CALCULATION:

We know that for a series LCR circuitthe resonating condition is given by:

⇒ XL = XC = X     -----(1)

If ϕ is the phase difference between the current and voltage in L-C-R circuit, then,

tanϕ=XLXCR     -----(2)

By equation 1 and equation 2,

tanϕ=XLXCR

tanϕ=XXR

⇒ tanϕ = 0

⇒ ϕ = 0

  • Hence, option 1 is correct.

In an LCR circuit at resonance

  1. The current and voltage are in same phase
  2. the impedance is maximum
  3. the current is minimum
  4. the current leads the voltage by π/2

Answer (Detailed Solution Below)

Option 1 : The current and voltage are in same phase

AC Voltage Applied to a Series LCR Circuit Question 15 Detailed Solution

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CONCEPT:

F1 P.Y 7.5.20 Pallavi D2

  • The ac circuit containing the capacitorresistor, and inductor is called an LCR circuit.
  • For a series LCR circuit, the total potential difference of the circuit is given by:

V=VR2+(VLVC)2

Where VR = potential difference across R, VL =  potential difference across L and VC =  potential difference across C

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

  • The resonant frequency of a series LCR circuit is given by

ν=12πLC

CALCULATION:

  • For a series LCR circuitImpedance (Z) of the circuit is given by:

Z=R2+(XLXC)2

  • Resonance is that condition when Inductive reactance is equal to capacitive reactance i.e. XL = XC.

Z=R2+(XLXL)2=R

  • So Impedance will be minimum i.e. equal to R
  • And hence the LCR circuit will behave like a normal circuit with resistance.
  • This means that in an LCR circuit at resonance the current and voltage are in phase
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