An a.c. source of frequency 50 Hz is connected to a 50 mH inductor and a bulb. The bulb glows with some brightness. Calculate the capacitance of the capacitor to be connected in series with the circuit so that the bulb glows with maximum brightness.

Solution

Given,
Frequency of AC source = 50 Hz
Inductance of bulb = 50 mH = 50

\times

10^-3 H

For, the bulb to glow with maximum brightness, it's impedence should be minimum.

The condition of minimum impedence is,

χ_{L} = χ_{C} \Rightarrow ωL = \frac{1}{ωC}

\Rightarrow

ω^{2} = \frac{1}{LC}

\Rightarrow

C = \frac{1}{ω^{2} L}

and,

ω = 2 π f

Therefore,
Capacitance of the capacitor, C =

\frac{1}{4 π^{2} f^{2} L}

= \frac{1}{4 \times 3.14 \times 3.14 \times 50 \times 50 \times 50 \times 10^{- 3}}

= 2

\times

10^-4 F

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