When (100 ~V) d.c. is applied across a coil, a current of 1 A flows through it. When (100 ~V) a.c. of (50 ~Hz) is applied to the same coil only 0.5 A flows. The inductance of the coil is

Correct answer: A. (0.55 ~H) For dc, V=I R ( R is resistance of inductor) 100=1 × R R=100 . For ac, Z= R^2+X_L^2 aligned & V=I Z & 100=0.5 R^2+X_L^2 & 100=0.5 (100)^2+X_L^2 aligned aligned & 100=(5)/(10) (100)^2+X_L^2 & X_L^2=(200)^2-(100)^2=300 × 100 & X_L= 3 ×(100)^2 =100 3 & ω L=100 3 2 π v L=100 3 & 2 × π × 50 × L=100 3 & 100 × π L=100 3 & L= 3 π = 3 3.14 L=0.55 H . aligned

AIIMS 2008

Exam: AIIMS
Year: 2008
Subject: Physics
Chapter: Alternating Current

AIIMS 2008 · Physics · Alternating Current

When (100 ~V) d.c. is applied across a coil, a current of 1 A flows...

When (100 \mathrm{~V}) d.c. is applied across a coil, a current of 1 A flows through it. When (100 \mathrm{~V}) a.c. of (50 \mathrm{~Hz}) is applied to the same coil only 0.5 A flows. The inductance of the coil is

Options

A(0.55 \mathrm{~H})
B(5.5 \mathrm{~mH})
C(0.55 \mathrm{~mH})
D(55 \mathrm{~mH}).

Correct answer

A. (0.55 \mathrm{~H})

Step-by-step solution

For dc,

V=I R

(

R

is resistance of inductor)

100=1 \times R \Rightarrow R=100 \Omega

. For ac,

Z=\sqrt{R^2+X_L^2}

\begin{aligned} & \because V=I Z \\ & \Rightarrow 100=0.5 \sqrt{R^2+X_L^2} \\ & \Rightarrow 100=0.5 \sqrt{(100)^2+X_L^2}\end{aligned}

\begin{aligned} & \Rightarrow 100=\frac{5}{10} \sqrt{(100)^2+X_L^2} \\ & X_L^2=(200)^2-(100)^2=300 \times 100 \\ & \Rightarrow X_L=\sqrt{3 \times(100)^2}=100 \sqrt{3} \\ & \Rightarrow \omega L=100 \sqrt{3} \Rightarrow 2 \pi v L=100 \sqrt{3} \\ & \Rightarrow 2 \times \pi \times 50 \times L=100 \sqrt{3} \\ & \Rightarrow 100 \times \pi L=100 \sqrt{3} \\ & \therefore L=\frac{\sqrt{3}}{\pi}=\frac{\sqrt{3}}{3.14} \Rightarrow L=0.55 \mathrm{H} .\end{aligned}

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