Electromagnetic Induction and AC

Question 1

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

Accepted Answer

7.8 A and 45°. For a series RLC circuit the impedance is Z = √(R² + (XL − XC)²) = √(20² + 20²) = 20√2 Ω, so I = 220/(20√2) ≈ 7.8 A. The phase angle follows from tan φ = (XL − XC)/R = 1, giving φ = 45°.

Answer

7.8 A and 30°

Answer

15.6 A and 30°

Answer

15.6 A and 45°

Question 2

Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is:

Accepted Answer

μ₀πn²r²l. Self-inductance is L = NΦ/i. Inside a long solenoid B = μ₀ni and the total turns are N = nl, so L = μ₀n²Al where A is the cross-sectional area. Substituting A = πr² gives L = μ₀πn²r²l.

Answer

μ₀n²r²l

Answer

(μ₀/2π)n²r²l

Answer

2μ₀πn²r²l

Question 3

An ac voltage V = 220 sin(2 × 10³ t) volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L = 10 mH, C = 25 μF, R = 100 Ω)

Accepted Answer

2.2 A. With ω = 2 × 10³ rad s⁻¹ the two reactances are XL = ωL = 20 Ω and XC = 1/(ωC) = 20 Ω. They are equal, so the circuit sits exactly at resonance, the impedance collapses to Z = R = 100 Ω, and the current amplitude is 220/100 = 2.2 A.

Answer

5.5 A

Answer

11.0 A

Answer

22.0 A

Q1.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

Q2.Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and μ₀ is the permeability of free space, the inductance of the solenoid is:

Q3.An ac voltage V = 220 sin(2 × 10³ t) volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given: L = 10 mH, C = 25 μF, R = 100 Ω)