CBSE Class 12 Physics Alternating Current Solved Examples

CBSE Class 12 Physics Alternating Current Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

SOLVED EXAMPLES: ALTERNATING CURRENT

Ex.1 What is average and r.m.s. current over half cycle if instantaneous current is given by
i = 4 sin ωt + 3 cos ωt.

Answer:
1. Given: i = 4 sin ωt + 3 cos ωt.
2. Peak current i₀ = √(4² + 3²) = √25 = 5 A.
3. Write i = 5 ( (4/5) sin ωt + (3/5) cos ωt ) = 5 sin (ωt + α), where cos α = 4/5 and sin α = 3/5.
4. Peak value i₀ = 5 A.
5. r.m.s. current i_rms = i₀/√2 = 5/√2 A = 3.54 A.
6. Average current over half cycle i_avg = 2i₀/π = 10/π A = 3.18 A.

Teacher's Note:
a) Combine the sine and cosine terms into one sinusoid using √(a² + b²) for the peak value.
b) The average over a half cycle is 2i₀/π, not i₀/2.
c) The r.m.s. value i₀/√2 does not depend on the phase angle α.

Ex.2 A 0.21 H inductor and a 12 Ω resistor are connected in series to a 20 V, 50 Hz ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

[Figure: Series circuit with a 12 Ω resistor and a 0.21 H inductor connected in series to an AC source marked 20 V, 50 Hz. Phasor diagram with V_R along the current direction, V_L at 90° to it, and the source voltage V_S at the angle φ from the current.]

Answer:
1. Given: R = 12 Ω, L = 0.21 H, V_rms = 20 V, f = 50 Hz.
2. Inductive reactance X_L = ωL = 2 × 3.14 × 50 × 0.21 = 65.94 Ω.
3. Impedance Z = √(R² + (ωL)²) = √(12² + 65.94²) = √4491.9 = 67 Ω.
4. Current I_rms = V_rms/Z = 20/67 = 0.30 A.
5. Phase angle φ = tan⁻¹(ωL/R) = tan⁻¹(65.94/12) = tan⁻¹(5.495) = 79.7°.
6. Hence, the current is 0.30 A and it lags the source voltage by 79.7°.

Teacher's Note:
a) The printed working used 220 V instead of the given 20 V, which gives a wrong current of 3.28 A; the correct current is 0.30 A.
b) The printed phase angle of 78.69° is an arithmetic slip; tan⁻¹(5.495) is 79.7°.
c) Check the impedance first, because the phase angle uses the same ratio ωL/R.

Ex.3 A resistance R, an inductance L and capacitor C are connected in series with an AC supply where R = 16Ω, inductive reactance X_L = 24Ω and capacitive reactance X_C = 12Ω. If the current in the circuit is 5 ampere find
(a) P.D. across R, L and C
(b) impedance of circuit,
(c) voltage of AC supply and
(d) phase angle.

[Figure: Series circuit with R, L and C in a row. Voltage across R is V_R, across C is V_C and across L is V_L, with the AC supply V_S across the whole series combination and current I flowing through all three.]

Answer:

(a) P.D. across R, L and C
1. V_R = I × R = 5 × 16 = 80 V.
2. V_L = I × X_L = 5 × 24 = 120 V.
3. V_C = I × X_C = 5 × 12 = 60 V.

(b) Impedance of circuit
1. Z = √(R² + (X_L - X_C)²) = √(16² + (24 - 12)²) = √(256 + 144) = √400 = 20 Ω.

(c) Voltage of AC supply
1. E = I × Z = 5 × 20 = 100 V.

(d) Phase angle
1. φ = tan⁻¹((X_L - X_C)/R) = tan⁻¹(12/16) = tan⁻¹(0.75) = 36.87°, that is 36°52′.

Teacher's Note:
a) The printed phase angle 36°46′ is wrong; tan⁻¹(0.75) is 36°52′.
b) Check that the three voltages V_R, V_L and V_C do not add arithmetically; the supply voltage is found from Z.
c) The phase angle is positive here, so the current lags the supply voltage because X_L is greater than X_C.

Ex.4 A current of 4 A flows in a coil when connected to a 12 V dc source. If the same coil is connected to a 12 V, 50 rad/s ac source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 μF condenser is connected in series with the coil.

Answer:
1. Resistance of the coil: R = V/I = 12/4 = 3 Ω. (The inductor has no effect on a dc source, so only R is measured.)
2. Impedance of the coil with ac: Z = 12/2.4 = 5 Ω.
3. Z² = R² + ω²L², so L = √(Z² - R²)/ω = √(25 - 9)/50 = 4/50 = 0.08 H.
4. Reactance of the inductor: X_L = ωL = 50 × 0.08 = 4 Ω.
5. Reactance of the capacitor: X_C = 1/(ωC) = 1/(50 × 2500 × 10⁻⁶) = 1/0.125 = 8 Ω.
6. Impedance with the capacitor in series: Z = √(R² + (X_L - X_C)²) = √(3² + (4 - 8)²) = √25 = 5 Ω.
7. Current I = V/Z = 12/5 = 2.4 A, and power factor cos φ = R/Z = 3/5 = 0.6.
8. Power developed P = V × I × cos φ = 12 × 2.4 × 0.6 = 17.28 W.

Teacher's Note:
a) A dc source gives the resistance only, because the inductor offers no reactance to direct current.
b) The power is taken from the resistance only, so P = I²R = 2.4² × 3 = 17.28 W gives the same answer.
c) Adding the capacitor makes the impedance equal to 5 Ω here, which is why the current stays 2.4 A.

Ex.5 A 100 V a.c source of frequency 500 Hz is connected to a series LCR circuit with L = 8.1 mH, C = 12.5 mF and R = 10 Ω. Find the potential difference across the resistance.

Answer:
1. Inductive reactance X_L = 2π × 500 × 8.1 × 10⁻³ = 25.45 Ω.
2. Capacitive reactance X_C = 1/(2π × 500 × 12.5 × 10⁻⁶) = 25.46 Ω.
3. Since X_L = X_C, the circuit is at resonance.
4. At resonance, V_L = V_C, so the impedance is Z = R and V = √(V_R² + (V_L - V_C)²) = V_R.
5. Hence, V_R = 100 V.

Teacher's Note:
a) The capacitance is 12.5 μF, not 12.5 mF; the printed reactance 25.478 Ω is only correct for 12.5 μF.
b) At resonance, the whole supply voltage appears across the resistance.
c) Resonance frequency 1/(2π√(LC)) is 500 Hz here, which confirms the condition.

Ex.6 Two inductances of 5.0 H and 10.0 H are connceted in parallel circuit. Find the equivalent inductance and r.m.s. voltage in each inductor and in mains circuit when connected to a source of 10 V AC.

[Figure: Two inductors of 5.0 H and 10.0 H connected in parallel across a 10 V AC source. Current I₁ flows through the 5.0 H inductor, current I₂ through the 10.0 H inductor, and the total current I is drawn from the mains.]

Answer:
1. For inductors in parallel: 1/L = 1/L₁ + 1/L₂ = 1/5 + 1/10 = 3/10.
2. Equivalent inductance L = 10/3 H = 3.33 H.
3. Assume the supply frequency is 50 Hz, as the solution uses, so ω = 2π × 50 rad/s.
4. The inductors are in parallel, so the r.m.s. voltage across each inductor is 10 V, the same as the mains voltage.
5. I_rms in L₁ = V/(ωL₁) = 10/(2π × 50 × 5) = 1/(50π) A.
6. I_rms in L₂ = V/(ωL₂) = 10/(2π × 50 × 10) = 1/(100π) A.
7. I_rms in the mains circuit = 1/(50π) + 1/(100π) = 3/(100π) A = 0.0095 A.

Teacher's Note:
a) The question does not give the frequency; the solution uses 50 Hz, which should be stated in the answer.
b) The current in the mains is the sum of the branch currents, not the sum of the inductances.
c) Check the mains current with the equivalent inductance: V/(ωL) = 10/(2π × 50 × 10/3) = 3/(100π) A, which agrees.

Ex.7 A 750 hertz, 20 V source is connected to a resistance of 100 ohm, an inductance of 0.1803 henry and a capacitance of 10 microfarad all in series. Calculate the time in which the resistance (thermal capacity = 2 joule / °C) will get heated by 10 °C ?

Answer:
1. Inductive reactance X_L = ωL = 2 × 3.14 × 750 × 0.1803 = 849.2 Ω.
2. Capacitive reactance X_C = 1/(ωC) = 1/(2 × 3.14 × 750 × 10 × 10⁻⁶) = 21.23 Ω.
3. Impedance Z = √(R² + (X_L - X_C)²) = √(100² + (849.2 - 21.23)²) = 834 Ω.
4. Power dissipated P = V_rms × I_rms × cos φ = V_rms² R/Z² = (20/834)² × 100 = 0.0574 W.
5. Heat required to raise the temperature by 10 °C: H = 2.0 × 10 = 20 J.
6. Using H = P × t, t = H/P = 20/0.0574 = 348 s.

Teacher's Note:
a) Power is dissipated only in the resistance, so use V_rms²R/Z².
b) The printed value 0.0575 W is a rounded figure; 0.0574 W gives t of about 348 s.
c) Heat is in joules, and the thermal capacity of 2 J/°C gives 20 J for a rise of 10 °C.

Ex.8 A series LCR circuit containing a resistance of 120 Ω has angular resonance frequency 4 × 10⁵ rad s⁻¹. At resonance, the voltage across resistance and inductance are 60 V and 40 V respectively. Find the values of L and C. At what frequency the current in the circuit lags the voltage by 45°?

[Figure: Phasor diagram with V_R along the current direction, V_L upward at 90° and V_C downward. The resultant of (V_L - V_C) and V_R gives the source voltage, which leads the current by 45°.]

Answer:

Values of L and C
1. For the resistance, V_R = I_rms R, so I_rms = 60/120 = 0.5 A.
2. For the inductor, V_L = I_rms ω₀L, so 40 = 0.5 × 4 × 10⁵ × L.
3. L = 40/(2 × 10⁵) = 2 × 10⁻⁴ H.
4. At resonance, ω₀L = 1/(ω₀C), so C = 1/(ω₀²L) = 1/((4 × 10⁵)² × 2 × 10⁻⁴) = 1/(3.2 × 10⁷) F = 1/32 μF = 0.03125 μF.

Frequency at which the current lags the voltage by 45°
1. The current lags the voltage by 45°, so tan 45° = (ωL - 1/(ωC))/R = 1, which gives ωL - 1/(ωC) = R.
2. Substitute 1/(ωC) = ω₀²L/ω, so ωL - ω₀²L/ω = R.
3. Multiply by ω: ω²L - ω₀²L = Rω, which gives 2 × 10⁻⁴ ω² - 120 ω - 2 × 10⁻⁴ × (4 × 10⁵)² = 0.
4. Divide by 2 × 10⁻⁴: ω² - 6 × 10⁵ ω - (4 × 10⁵)² = 0.
5. Solve: ω = (6 × 10⁵ ± √(36 × 10¹⁰ + 64 × 10¹⁰))/2 = (6 × 10⁵ ± 10 × 10⁵)/2, so ω = 8 × 10⁵ or -2 × 10⁵.
6. Reject the negative root. Hence ω = 8 × 10⁵ rad/s, which is a frequency of about 1.27 × 10⁵ Hz.

Teacher's Note:
a) The printed step "ω²L - ω₀²L = 2ω" should read "Rω = 120ω", because the resistance is 120 Ω.
b) The answer is an angular frequency, so the unit is rad/s, not Hz.
c) Use the resonance condition to eliminate C before forming the quadratic equation.

Ex.9 A LCR circuit has L = 10 mH, R = 3 ohm and C = 1μF connected in series to a source of 15 cos (ωt) volt. Calculate the current amplitude and the average power dissipated per cycle at a frequency that is 10% lower than the resonant frequency.

Answer:
1. Resonant frequency ω_R = 1/√(LC) = 1/√(10 × 10⁻³ × 1 × 10⁻⁶) = 1/√(10⁻⁸) = 10⁴ rad/s.
2. Frequency 10% lower: ω = 10⁴ - 10⁴ × 10/100 = 9 × 10³ rad/s.
3. X_L = ωL = 9 × 10³ × 10 × 10⁻³ = 90 Ω.
4. X_C = 1/(ωC) = 1/(9 × 10³ × 1 × 10⁻⁶) = 111.11 Ω.
5. Z = √(R² + (X_L - X_C)²) = √(3² + (90 - 111.11)²) = √454.5 = 21.32 Ω.
6. Current amplitude I₀ = E₀/Z = 15/21.32 = 0.704 A.
7. Power factor cos φ = R/Z = 3/21.32 = 0.141.
8. Average power P = ½ E₀ I₀ cos φ = ½ × 15 × 0.704 × 0.141 = 0.742 W.

Teacher's Note:
a) The printed average power 0.744 W comes from rounding the intermediate values; the value from unrounded numbers is 0.742 W.
b) Use the peak voltage E₀ = 15 V with the factor ½ for average power.
c) At a frequency below resonance the circuit is capacitive, so X_C is greater than X_L.