An electric bulb has a rated power of 50 ~W at 100 ~V . If it is used on an AC source of 200 ~V , 50 ~Hz , a choke has to be used in series with it. This choke should have an inductance of

Correct answer: D. 1.1 H Here, P=50 ~W, V=100 ~VI=(P)/(V)=(50)/(100)=0.5 ~A, R=(V)/(I)=(100)/(0.5)=200 Let L be the inductance of the choke coil aligned & I_v=(E_v)/(Z) or Z=(E_v)/(I_v)=(200)/(0.5)=400 & =100 12 & aligned Now, X_L= Z^2-R^2 = 400^2-200^2 =100 12 or, ω L=100 × 2 3 aligned L= 200 3 ω = 200 3 2 π v = 200 3 100 π & =(2 × 1.732)/(3.14) & =1.1 H aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Alternating Current

AIIMS 2018 · 26 May · Physics · Alternating Current

An electric bulb has a rated power of 50 ~W at 100 ~V . If it is...

An electric bulb has a rated power of

50 \mathrm{~W}

100 \mathrm{~V}

. If it is used on an AC source of

200 \mathrm{~V}

50 \mathrm{~Hz}

, a choke has to be used in series with it. This choke should have an inductance of

Options

A $1 \mathrm{mH}$
B $0.1 \mathrm{mH}$
C $0.1 \mathrm{H}$
D $1.1 \mathrm{H}$

Correct answer

D. $1.1 \mathrm{H}$

Step-by-step solution

Here,

P=50 \mathrm{~W}, V=100 \mathrm{~V}I=\frac{P}{V}=\frac{50}{100}=0.5 \mathrm{~A}, R=\frac{V}{I}=\frac{100}{0.5}=200 \Omega

Let

L

be the inductance of the choke coil

\begin{aligned}& \therefore \quad I_v=\frac{E_v}{Z} \text{ or }Z=\frac{E_v}{I_v}=\frac{200}{0.5}=400 \Omega \\ & =100 \sqrt{12} \Omega\\ & \end{aligned}

Now,

X_L=\sqrt{Z^2-R^2}=\sqrt{400^2-200^2}=100 \sqrt{12} \Omega

or,

\omega L=100 \times 2 \sqrt{3} \Omega

\begin{aligned} \Rightarrow L=\frac{200 \sqrt{3}}{\omega}=\frac{200 \sqrt{3}}{2 \pi v}=\frac{200 \sqrt{3}}{100 \pi} & =\frac{2 \times 1.732}{3.14} \\ & =1.1 \mathrm{H}\end{aligned}

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