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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Alternating Current

A bulb connected to 50 V, DC consumes 20 W power. Then the bulb is connected to a capacitor in an a.c. power supply of 250 V, 50 Hz. Find the value of the capacitor required so that the bulb draws the same amount of current.

(A) $1.891 \mu F$ (B) $2.649 \mu F$

(C) $3.296 \mu F$

(D) $5.198 \mu F$

1 Answer

$\text{Power} \;P = V \times I \rightarrow I = \large\frac{P}{V} $$ = \large\frac{20W}{50V} $$ = 0.4A$
$\Rightarrow \text{ Resistance}\; R = \large\frac{V}{I} $$ = \large\frac{50V}{0.4A} $$ = 125 \Omega$
$\Rightarrow \text{Impedence} Z = \large\frac{V_{\text{power supply}}}{I}$$ = \large\frac{250V}{0.4A} $$ = 625 \Omega$
$\text{Impedence} \; Z = \large\sqrt ( R^2 + ( \large\frac{1}{\omega C})^2)$
$\text{Impedence} \; Z = \large\sqrt ( R^2 + ( \large\frac{1}{2 \pi V C})^2)$
$\Rightarrow Z^2 = R^2 + ( \large\frac{1}{2\pi V C})^2$
Re-arranging and solving for C, we get: $C = \large\frac{1}{2\pi\sqrt (Z^2 - R^2)}$$ = \large\frac{1}{2\pi \times 50 \sqrt(625^2 - 125^2)}$
$\Rightarrow C = \large\frac{1}{2\pi \times 50 \times 612.87}$$ \approx 5.198 \mu F$
answered Mar 12, 2014 by balaji.thirumalai
 

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