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Learn/ 12th Class Physics/ Unit 17 /Numericals

Unit 17: Alternating Current — Numericals

12th Class Physics · Unit 17: Alternating Current

Num17.1.AC circuit configured by a resistance of 20 Ω, capacitor of capacitance 40 μF is connected to an AC supply of 110 V having frequency 50 Hz. Calculate: (a) capacitive reactance (b) current in the circuit (c) suggest phase between the voltage and current in a capacitive circuit.
Num17.2.An alternating current I is represented by the equation: I = 420 sin(100πt). Find (a) the rms current (b) the frequency.

Given I = 420 sin(100πt) A, comparing with I = I₀ sin(ωt): I₀ = 420 A, ω = 100π rad/s.

(a) Irms = I₀/√2 = 420/1.4142 = 296.98 A ≈ 296.94 A.

(b) ω = 2πf ⟹ f = ω/(2π) = 100π/(2π) = 50 Hz.

Final (a) Irms ≈ 296.94 A (b) f = 50 Hz.

Num17.3.A 50 Hz AC of peak value 1 A flows through primary coil of a transformer. If the mutual inductance between the primary and secondary be 1.5 H, then find the mean value of induced voltage.
Num17.4.A 220 V AC source of frequency 2 kHz is connected across a capacitor of 10 μF. Find out: (a) the reactance X_C of capacitor (b) the current in the circuit.

Given Vrms = 220 V, f = 2 kHz = 2000 Hz, C = 10 μF = 10×10⁻⁶ F.

(a) XC = 1/(2πfC) = 1/(2π × 2000 × 10×10⁻⁶) = 1/0.12566 = 7.96 Ω.

(b) Irms = Vrms/XC = 220/7.96 = 27.65 A ≈ 27.63 A.

Final (a) XC ≈ 7.96 Ω (b) Irms ≈ 27.63 A.

Num17.5.In a coil of inductance 150 mH, a current changes steadily from 50 mA to 30 mA in 164 ms, find: (a) the magnitude (b) direction of induced emf.

Given L = 150 mH = 0.15 H, Iinitial = 50 mA, Ifinal = 30 mA, so ΔI = 50 − 30 = 20 mA = 0.02 A, Δt = 164 ms = 0.164 s.

(a) |ε| = L(ΔI/Δt) = 0.15 × (0.02/0.164) = 0.15 × 0.1220 = 0.0183 V = 18.3 mV.

(b) Since the current is decreasing, by Lenz's law the induced emf opposes this decrease, i.e., it acts in the same direction as the original current, trying to maintain it.

Final (a) ε ≈ 18.29 mV (b) Induced emf acts in the same direction as the original current.

Num17.6.A choke coil is needed to operate a lamp at 160 V_rms and 50 Hz. The lamp has a resistance of 5 Ω when running at 10 A_rms. Calculate the inductance of the choke coil.

Given Vrms = 160 V, f = 50 Hz, R (lamp) = 5 Ω, Irms = 10 A. The choke (negligible resistance, inductance L) is in series with the lamp.

Total impedance Z = Vrms/Irms = 160/10 = 16 Ω.

Since Z = √(R² + XL²): XL² = Z² − R² = 16² − 5² = 256 − 25 = 231, so XL = √231 = 15.20 Ω.

Inductance L = XL/(2πf) = 15.20/(2π × 50) = 15.20/314.16 = 0.0484 H = 48.4 mH.

Final L ≈ 48.4 mH.