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An electric bulb rated for $\mathrm{500 \mathrm{~W} \: at \: 100 \mathrm{~V}}$ is used in a circuit having $\mathrm{200 \mathrm{~V}}$ supply. Find the resistance $\mathrm{R}$ that must be put in series with the buck, so that the bulb draws $\mathrm{500 \mathrm{~W}}$ .

Option: 1
$40 \Omega$

Option: 2
$60 \Omega$

Option: 3
$\mathrm{52\Omega }$

Option: 4
$20 \Omega$

Answers (1)

best_answer

current needed for full power consumption

$\mathrm{i=\frac{p}{v}=\frac{500}{100}=5 \mathrm{~A}}$

and resistance of bulb, $\mathrm{R_0=\frac{v^2}{\rho}=\frac{(100)^2}{500}=20\Omega}$

If $\mathrm{R}$ is required resistance, then

$\mathrm{ i=\frac{V_{\text {supply }}}{R_0+R} }$

$\mathrm{ i=\frac{200}{20+R} }$

but $\mathrm{ i=5 \mathrm{~A}, }$

$\mathrm{ 5=\frac{200}{20+R} }$

$\mathrm{ 100+5 R =200 }$

$\mathrm{ 5 R =100 }$

$\mathrm{ R =20\Omega }$

Hence option 4 is correct.

Posted by

shivangi.shekhar

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