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

Correct option is: (2) 7.8 A and 45°

X_L = 45 W, X_C = 25 W, R = 20 W

⇒ I = 220 / √((X_L − X_C)² + R²) = 220 / √((45 − 25)² + 20²)

⇒ = 220 / (2√2) = 11 / √2 = 7.779 A

⇒ tan φ = (X_L − X_C) / R = (45 − 25) / 20 = 1

⇒ φ = 45°

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 X_L and total impedance Z are:

X_L = ωL = 2πνL = 1.88 Ω

Z = √(X_L² + R²) = 2.75 Ω

Thus,

I = V / Z ≠ 0

V₁ = X_LI = 1.88I

V₂ = RI = 2I

V₂ > V₁

In circuit (s): Inductive reactance X_L, capacitive reactance X_C, and impedance Z are:

X_L = 1.88 Ω

X_C = 1 / (ωC) = 1061 Ω

Z = X_C − X_L = 1059 Ω

So the current is:

I = V / Z ≠ 0

V₁ = X_LI = 1.88I

V₂ = X_CI = 1061I

V₂ > V₁

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

Calculation:

\(Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})}\)

\(=\sqrt{(300)^2+\left(1000\times 0.9-\dfrac{1}{1000\times 2\times 10^{-6}}\right)^2}\)

\(Z=\sqrt{(300)^2+(400)^2}\Rightarrow 500\Omega \)

The correct option is (a)

Concept:

AC Circuit with Impedances: The circuit consists of a capacitor, inductor, and resistors in two branches. The total impedance is calculated separately for each branch.

• Impedance of a Capacitor: Z_C = 1 / (ωC)
• Impedance of an Inductor: Z_L = ωL
• Net Impedance of a Branch: Z = √(R² + X²), where X is the reactance.

Calculation:

Given,

Capacitance, C = 100 μF = 100 × 10^-6 F

Inductance, L = 0.5 H

Resistances: R₁ = 100 Ω, R₂ = 50 Ω

AC voltage, V = 20 V

Angular frequency, ω = 100 rad/s

⇒ Net impedance in the upper branch:

Z₁ = √((1 / (ωC))² + R₁²)

⇒ Z₁ = √((1 / (100 × 100 × 10^-6))² + 100²)

⇒ Z₁ = √(100² + 100²) = 100√2 Ω

⇒ Current in the upper branch:

I₁ = V / Z₁ = 20 / (100√2)

⇒ I₁ = 0.2 / √2 A

⇒ Net impedance in the lower branch:

Z₂ = √((ωL)² + R₂²)

⇒ Z₂ = √((0.5 × 100)² + 50²)

⇒ Z₂ = √(50² + 50²) = 50√2 Ω

⇒ Current in the lower branch:

I₂ = V / Z₂ = 20 / (50√2)

⇒ I₂ = 0.4 / √2 A

⇒ Total current in the circuit:

I = √(I₁² + I₂²) (as they are 90° out of phase)

⇒ I = √((0.2 / √2)² + (0.4 / √2)²)

⇒ I = √(0.04 + 0.16) = √0.2 = 0.3 A

⇒ Voltage across 100Ω resistor:

V_R1 = I₁ × R₁ = (0.2 / √2) × 100

⇒ V_R1 = 10√2 V

Voltage acroos 50 Ω is 

V₅₀ =  I₂ 50 Ω = 10√2 V 

Concept:

LCR Circuit Behavior:

• In a series LCR circuit, the voltage and current are related by the impedance of the circuit, which depends on the values of the inductance (L), capacitance (C), and resistance (R).
• The inductive reactance (XL) and capacitive reactance (Xc) are given by the formulas:
  XL = 2πfL, Xc = 1 / (2πfC),
  where f is the frequency of the source.
• When the current leads the voltage in an LCR circuit, the net reactance is capacitive, i.e., the capacitive reactance Xc is greater than the inductive reactance Xl.
• Thus, if the current leads the source voltage, the circuit is capacitive in nature, and Xc > XL.

Calculation/Explanation:

We are given that the current leads the source voltage in the LCR circuit. This situation arises when the capacitive reactance (Xc) is greater than the inductive reactance (XL), making the circuit behave like a capacitor.

From the behavior of the LCR circuit:

• Option 1: Xc > XL – This is the correct condition when the current leads the voltage, as it indicates a net capacitive reactance.
• Option 2: XL > Xc – This would occur if the current lagged the voltage, indicating an inductive circuit.
• Option 3: XL = Xc ≠ 0 – This would result in a purely resistive circuit where the current is in phase with the voltage.
• Option 4: XL = Xc = 0 – This would correspond to a purely resistive circuit with no reactance at all.

∴ The correct answer is Option 1: Xc > XL.

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

CALCULATION:

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

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

• Inductive reactance,

⇒ XL = Lω  

• Capacitive reactance

​\(\Rightarrow X_c=\frac{1}{C\omega}\)

• Resonance will take place when XL = X_C.

​⇒ XL = X_C

\(\Rightarrow L\omega =\frac{1}{C\omega}\)

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
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, X_c = X_L, then the value of impedance is given by

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

\(\Rightarrow Z = \sqrt{R^{2}+0}\)

\(\Rightarrow Z = R\)

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

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

CALCULATION:

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

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

• Inductive reactance,

⇒ XL = Lω  

• Capacitive reactance

​\(\Rightarrow X_c=\frac{1}{Cω}\)

• Resonance will take place when XL = XC.

​⇒ XL = XC

\(\Rightarrow Lω =\frac{1}{Cω}\)

\(\Rightarrow ω =\frac{1}{\sqrt{LC}}\)

As we know, ω = 2πf

Where f = frequency

\(\Rightarrow f =\frac{1}{2\pi\sqrt{LC}}\)

Given:

\(R=30\Omega,X_L=200\Omega,X_C=160\Omega,V=220V\)

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,

• \(Z=\sqrt{R^2+(X_L-X_C)^2}\)

Where \(R\) is Resistance,

 \(X_L\) is inductive reactance,

\(X_C\) is capacitive reactance

Explanation:

• \(Z=\sqrt{R^2+(X_L-X_C)^2}\)
• \(Z=\sqrt{30^2+(200-160)^2}=\sqrt{30^2+40^2}=\sqrt{2500}=50\Omega\)
• \(Z=\sqrt{30^2+40^2}=\sqrt{2500}=50\Omega\)

Current, I\(=\frac{V}{Z}=\frac{220}{50}=\frac{22}{5}A\)

Potential drop across the resistor is \(V_d=IR=\frac{22}{5}\times30=132V\)

The correct answer is Option-1-132V.

CONCEPT:

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

\(⇒ V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(⇒ Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

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

\(⇒ f = \frac{1}{{2\pi }}\sqrt {\frac{1}{{LC}}}\)

CALCULATION:

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

\(⇒ f = {\frac{1}{2\pi\sqrt{LC}}}\)

• 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.   

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

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

\(\Rightarrow \nu =\frac{1}{2\pi\sqrt{LC}}\)

​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 (X_C) (Ohms is the unit)

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

​CALCULATION:

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

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

• For resonating condition,

\(\Rightarrow X_{L}=X_{C}\)

\(\Rightarrow Z= R\)

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

CONCEPT:

LCR CIRCUIT:

•  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:

\(⇒ V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(⇒ Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =inductive reactance and XC = capacitive reactive

• At resonance X_c = X_L,  the frequency of oscillation is given by

\(⇒ \nu = \frac{1}{2\pi\sqrt{LC}}\)

Where L = Inductance, C = Capacitance

EXPLANATION :

• The power factor of LCR is given by

⇒ P  = VI cosϕ  ----(1)

\(⇒ Cosϕ = \frac{R}{Z}\)

at resonance R = Z

\(⇒ Cosϕ = \frac{Z}{Z} = 1\)

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.

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:

 \(X_L = ω L=2\pi fL\)

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,.

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

\(Z = \sqrt{R^2 + {ω L}^2}\)

where R is resistance and ωL is Inductance reactance.

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

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

\(⇒ \nu =\frac{1}{2\pi\sqrt{LC}}\)

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

\(⇒ tanϕ=\frac{X_L-X_C}{R}=\frac{V_L-V_C}{V_R}\)

​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 circuit, the resonating condition is given by:

⇒ X_L = X_C = X     -----(1)

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

\(⇒ tanϕ=\frac{X_L-X_C}{R}\)     -----(2)

By equation 1 and equation 2,

\(⇒ tanϕ=\frac{X_L-X_C}{R}\)

\(⇒ tanϕ=\frac{X-X}{R}\)

⇒ tanϕ = 0

⇒ ϕ = 0

• Hence, option 1 is correct.

CONCEPT:

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

\(V = \sqrt {{V_R^2} + {{\left( {{V_L} - {V_C}} \right)}^2}} \)

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

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

\(\)\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
Where R = resistance, XL =induvtive reactance and XC = capacitive reactive

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

\(\Rightarrow \nu =\frac{1}{2\pi\sqrt{LC}}\)

CALCULATION:

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

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

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

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

• 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