Chapter 7: Alternating Current Application Questions | physics Class 12 CBSE

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Chapter 7: Alternating Current Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A 100 W light bulb is rated for 220 V ac supply. Calculate (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb.

The resistance of the bulb: R = V²/P = (220)²/100 = 48400/100 = 484 Ω. This uses the rms voltage directly since the rating is given in rms values.
The peak voltage: vm = √2 × V = 1.414 × 220 ≈ 311 V. This is the maximum instantaneous voltage the bulb experiences during each cycle.
The rms current: I = P/V = 100/220 ≈ 0.454 A. Alternatively, I = V/R = 220/484 = 0.454 A. The bulb dissipates real power because it is a purely resistive load.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A pure inductor of 25.0 mH is connected to a 220 V, 50 Hz ac source. Find the inductive reactance and the rms current in the circuit. What would happen to the current if frequency is doubled?

Inductive reactance: XL = 2πνL = 2 × 3.14 × 50 × 25 × 10⁻³ = 7.85 Ω. This represents the opposition offered by the inductor at 50 Hz.
Rms current: I = V/XL = 220/7.85 ≈ 28 A. The inductor allows a large current at 50 Hz because its reactance is small at this frequency.
If frequency is doubled to 100 Hz: XL doubles to 15.7 Ω, and the rms current halves to approximately 14 A. This demonstrates that inductive reactance increases linearly with frequency, reducing current.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms current. If frequency is doubled, what happens to the reactance and current?

Capacitive reactance: XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 15 × 10⁻⁶) = 212 Ω. This is the opposition offered by the capacitor at 50 Hz.
Rms current: I = V/XC = 220/212 ≈ 1.04 A. Peak current: im = √2 × 1.04 ≈ 1.47 A. The current oscillates between +1.47 A and −1.47 A and leads voltage by π/2.
If frequency is doubled to 100 Hz: XC halves to 106 Ω, and the current doubles to approximately 2.08 A. This shows capacitive reactance is inversely proportional to frequency — higher frequency means easier passage of ac through a capacitor.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A resistor of 200 Ω and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source. Calculate the current in the circuit and the rms voltage across R and C. Verify that their algebraic sum exceeds 220 V and resolve the paradox.

Impedance: Z = √(R² + XC²) = √(200² + 212.3²) = √(40000 + 45071) = √85071 ≈ 291.7 Ω. Current: I = V/Z = 220/291.7 ≈ 0.755 A.
VR = IR = 0.755 × 200 = 151 V; VC = IXC = 0.755 × 212.3 ≈ 160.3 V. Algebraic sum = 151 + 160.3 = 311.3 V > 220 V. This seems paradoxical.
The paradox resolves because VR and VC are 90° out of phase and cannot be added algebraically. Using the Pythagorean theorem: V = √(VR² + VC²) = √(151² + 160.3²) = √(22801 + 25696) ≈ 220 V, consistent with the source.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

With reference to a series LCR circuit with R = 3 Ω, L = 25.48 mH, C = 796 μF connected to a 283 V peak, 50 Hz source, calculate (a) impedance, (b) phase difference, and (c) power dissipated.

XL = 2π × 50 × 25.48 × 10⁻³ = 8 Ω; XC = 1/(2π × 50 × 796 × 10⁻⁶) = 4 Ω. Impedance: Z = √(3² + (8−4)²) = √(9+16) = 5 Ω.
Phase difference: φ = tan⁻¹[(XC − XL)/R] = tan⁻¹[(4−8)/3] = tan⁻¹(−4/3) = −53.1°. Negative φ means current lags voltage, so circuit is inductive (XL > XC).
Rms current: I = (vm/√2)/Z = (283/√2)/5 = 40 A. Power dissipated: P = I²R = (40)² × 3 = 4800 W. Power factor = cos(−53.1°) = 0.6.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

For the same LCR circuit (R = 3 Ω, L = 25.48 mH, C = 796 μF, vm = 283 V), suppose the source frequency is varied. Find the resonant frequency and calculate the current and power at resonance.

Resonant frequency: ω₀ = 1/√(LC) = 1/√(25.48 × 10⁻³ × 796 × 10⁻⁶) = 222.1 rad/s; ν₀ = ω₀/(2π) ≈ 35.4 Hz.
At resonance, Z = R = 3 Ω (minimum impedance). Rms current: I = V/R = (283/√2)/3 = 66.7 A — maximum current condition.
Power dissipated at resonance: P = I²R = (66.7)² × 3 ≈ 13.35 kW. This is much larger than the 4800 W dissipated at 50 Hz, confirming resonance maximizes power transfer to the resistive element.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A series LCR circuit has R = 20 Ω, L = 1.5 H, C = 35 μF connected to a variable-frequency 200 V ac supply. When the source frequency equals the natural frequency, calculate the average power transferred in one cycle.

At resonance (source frequency = natural frequency), impedance Z = R = 20 Ω (minimum) and power factor cos φ = 1.
Rms current at resonance: I = V/R = 200/20 = 10 A. This is the maximum possible current for this circuit.
Average power at resonance: P = I²R = (10)² × 20 = 2000 W = 2 kW. Since cos φ = 1 at resonance, all apparent power becomes real power, and it is entirely dissipated in R.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 8

Applied Concept

A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. Find (a) the rms current, (b) the net power consumed over a full cycle, and (c) the peak voltage across the resistor.

Rms current: I = V/R = 220/100 = 2.2 A. In a purely resistive circuit, voltage and current are in phase and Ohm's law directly gives the rms current.
Net power consumed: P = I²R = (2.2)² × 100 = 484 W. Alternatively, P = V²/R = 220²/100 = 484 W. Power factor = 1 for a pure resistor.
Peak voltage across resistor: vm = √2 × V = 1.414 × 220 ≈ 311 V. This equals the peak source voltage since V = IR holds for rms values and correspondingly vm = im R for peak values.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

In a power transmission scenario, 1000 W is to be delivered at 220 V through wires of total resistance 10 Ω. Compare power loss if delivered (a) directly at 220 V and (b) stepped up to 2200 V and then stepped down.

At 220 V: I = P/V = 1000/220 ≈ 4.55 A. Power loss in wires = I²R = (4.55)² × 10 = 206.9 W — about 20.7% of transmitted power is wasted.
At 2200 V (10× step-up): I = P/V = 1000/2200 ≈ 0.455 A. Power loss = I²R = (0.455)² × 10 = 2.07 W — only 0.21% is wasted. Loss reduces by factor of 100.
This demonstrates why high-voltage transmission is essential: stepping up voltage by a factor of 10 reduces current by 10, and since loss scales as I², the power loss reduces by a factor of 100, making long-distance transmission economically viable.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

A charged 30 μF capacitor is connected to a 27 mH inductor. With reference to the principle of an LC circuit, find the angular frequency of free oscillations.

The angular frequency of free oscillations in an LC circuit is given by ω₀ = 1/√(LC), where energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.
Substituting values: ω₀ = 1/√(27 × 10⁻³ × 30 × 10⁻⁶) = 1/√(810 × 10⁻⁹) = 1/(28.46 × 10⁻⁶) ≈ 1.11 × 10³ rad/s.
This frequency is analogous to the resonant frequency of a series RLC circuit (with R = 0). It represents the natural frequency at which charge sloshes back and forth between the capacitor and inductor without any external driving source.

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Chapter sections

Chapter 7: Alternating Current | Application Questions

A 100 W light bulb is rated for 220 V ac supply. Calculate (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb.

A pure inductor of 25.0 mH is connected to a 220 V, 50 Hz ac source. Find the inductive reactance and the rms current in the circuit. What would happen to the current if frequency is doubled?

A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the rms current. If frequency is doubled, what happens to the reactance and current?

A resistor of 200 Ω and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source. Calculate the current in the circuit and the rms voltage across R and C. Verify that their algebraic sum exceeds 220 V and resolve the paradox.

With reference to a series LCR circuit with R = 3 Ω, L = 25.48 mH, C = 796 μF connected to a 283 V peak, 50 Hz source, calculate (a) impedance, (b) phase difference, and (c) power dissipated.

For the same LCR circuit (R = 3 Ω, L = 25.48 mH, C = 796 μF, vm = 283 V), suppose the source frequency is varied. Find the resonant frequency and calculate the current and power at resonance.

A series LCR circuit has R = 20 Ω, L = 1.5 H, C = 35 μF connected to a variable-frequency 200 V ac supply. When the source frequency equals the natural frequency, calculate the average power transferred in one cycle.

A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. Find (a) the rms current, (b) the net power consumed over a full cycle, and (c) the peak voltage across the resistor.

In a power transmission scenario, 1000 W is to be delivered at 220 V through wires of total resistance 10 Ω. Compare power loss if delivered (a) directly at 220 V and (b) stepped up to 2200 V and then stepped down.

A charged 30 μF capacitor is connected to a 27 mH inductor. With reference to the principle of an LC circuit, find the angular frequency of free oscillations.

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