An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 Ω. The phase difference between current and voltage is

Question

An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 Ω. The phase difference between current and voltage is (A)  90° (B)  60° (C)  75° (D) 45°

Tardigrade Admin · Accepted Answer

tan φ = ( XL/ R) = ( ω L/ R) = ( 2 π f L/ R ) ∴ tan φ = ( 2 × 3.14 × 50 × 0.5 / 157 ) = 1 ∴ φ = 45°

Q. An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 $Ω$ . The phase difference between current and voltage is

$9 0^{\circ}$

$6 0^{\circ}$

$7 5^{\circ}$

$4 5^{\circ}$

$3 0^{\circ}$

$tan ϕ = \frac{X _{L}}{R} = \frac{ω L}{R} = \frac{2 π f L}{R}$
$∴ tan ϕ = \frac{2 \times 3.14 \times 50 \times 0.5}{157} = 1$
$∴ ϕ = 4 5^{\circ}$

Q. An AC source of frequency 50 Hz is connected in series to an inductance of 0.5 H and resistance of 157 Ω. The phase difference between current and voltage is

90∘

60∘

75∘

45∘

30∘

tanϕ=RXL​​=RωL​=R2πfL​ ∴tanϕ=1572×3.14×50×0.5​=1 ∴ϕ=45∘