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 __________.
Answer (integer)
400
Solution
Given, power factor of LR circuit,
<br><br>cos $\phi$ = 0.8 = ${R \over {\sqrt {{R^2} + X_L^2} }}$ = ${R \over Z}$
<br><br>We know,
<br>Power, P = ${{V_{rms}^2} \over {{Z^2}}} \times R$
<br><br>$\Rightarrow$ 400 = ${{{{\left( {250} \right)}^2} \times 0.8Z} \over {{Z^2}}}$
<br><br>$\Rightarrow$ Z = 125
<br><br>$\therefore$ R = 0.8 $\times$ 125 = 100 $\Omega$
<br><br>As Z<sup>2</sup> = $X_L^2 + {R^2}$
<br><br>$\Rightarrow$ ${\left( {125} \right)^2} = X_L^2 + {\left( {100} \right)^2}$
<br><br>$\Rightarrow$ X<sub>L</sub> = 75
<br><br>In 2<sup>nd</sup> case given.
<br><br>Power factor = 1
<br><br>that means
X<sub>L</sub>
= X<sub>C</sub>
(Resonance condition)
<br><br>X<sub>L</sub> = ${1 \over {{\omega _c}}}$
<br><br>$\Rightarrow$ 75 = ${1 \over {\left( {2\pi F} \right)C}}$
<br><br>$\Rightarrow$ C = ${1 \over {\left( {2\pi \times 50} \right)75}}$
<br><br>Also given, C = $\left( {{n \over {3\pi }}} \right)$ $\mu$F
<br><br>$\therefore$ $${1 \over {\left( {2\pi \times 50} \right)75}} = {{n \times {{10}^{ - 6}}} \over {3\pi }}$$
<br><br>$\Rightarrow$ n = 400
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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