Question 127: AC fundamentals

Question

In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor
of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in
series to the L and R. Taking the value of C as $$\left( {{n \over {3\pi }}} \right)$$ $$\mu $$F,  then value of n is __________.

Accepted Answer

Given, power factor of LR circuit,
cos $$\phi $$ = 0.8 = $${R \over {\sqrt {{R^2} + X_L^2} }}$$ = $${R \over Z}$$
We know,
Power, P = $${{V_{rms}^2} \over {{Z^2}}} 	imes R$$
$$ \Rightarrow $$ 400 = $${{{{\left( {250} ight)}^2} 	imes 0.8Z} \over {{Z^2}}}$$
$$ \Rightarrow $$ Z = 125
$$ 	herefore $$ R = 0.8 $$ 	imes $$ 125 = 100 $$\Omega $$
As Z
2
= $$X_L^2 + {R^2}$$
$$ \Rightarrow $$ $${\left( {125} ight)^2} = X_L^2 + {\left( {100} ight)^2}$$
$$ \Rightarrow $$ X
L
= 75
In 2
nd
case given.
Power factor = 1
that means
X
L
= X
C
(Resonance condition)
X
L
= $${1 \over {{\omega _c}}}$$
$$ \Rightarrow $$ 75 = $${1 \over {\left( {2\pi F} ight)C}}$$
$$ \Rightarrow $$ C = $${1 \over {\left( {2\pi  	imes 50} ight)75}}$$
Also given, C = $$\left( {{n \over {3\pi }}} ight)$$ $$\mu $$F
$$ 	herefore $$ $${1 \over {\left( {2\pi  	imes 50} ight)75}} = {{n 	imes {{10}^{ - 6}}} \over {3\pi }}$$
$$ \Rightarrow $$ n = 400

AC fundamentals

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