A coil of resistance 300 Ω and inductance 1 H is connected across an alternating voltage of frequency 300 / 2 π Hz. Calculate the phase difference between voltage and current in the circuit.

Question

A coil of resistance 300 Ω and inductance 1 H is connected across an alternating voltage of frequency 300 / 2 π Hz. Calculate the phase difference between voltage and current in the circuit. (A) 60° (B) 30° (C) 45° (D) 90°

Tardigrade Admin · Accepted Answer

R prime=300 Ω, L =1 H , v=(300/2 π) Hz ∴ tan φ=(XL/R) ⇒ (2 π v L/R) ∴ tan φ=(2 π × (300/2 π) × 1/300)=1 φ=450

Q. A coil of resistance $300Ω$ and inductance $1 H$ is connected across an alternating voltage of frequency $300/2 π Hz$ . Calculate the phase difference between voltage and current in the circuit.

$6 0^{\circ}$

$3 0^{\circ}$

$4 5^{\circ}$

$9 0^{\circ}$

$R^{'} = 300Ω, L = 1 H, v = \frac{300}{2 π} Hz$
$∴ tan ϕ = \frac{X _{L}}{R}$
$\Rightarrow \frac{2 π vL}{R}$
$∴ tan ϕ = \frac{2 π \times \frac{300}{2 π} \times 1}{300} = 1$
$ϕ = 4 5^{0}$

Q. A coil of resistance 300Ω and inductance 1H is connected across an alternating voltage of frequency 300/2πHz. Calculate the phase difference between voltage and current in the circuit.

60∘

30∘

45∘

90∘