JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 14)
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
400
Explanation
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 = $${{V_{rms}^2} \over {{Z^2}}} \times R$$
$$ \Rightarrow $$ 400 = $${{{{\left( {250} \right)}^2} \times 0.8Z} \over {{Z^2}}}$$
$$ \Rightarrow $$ Z = 125
$$ \therefore $$ R = 0.8 $$ \times $$ 125 = 100 $$\Omega $$
As Z2 = $$X_L^2 + {R^2}$$
$$ \Rightarrow $$ $${\left( {125} \right)^2} = X_L^2 + {\left( {100} \right)^2}$$
$$ \Rightarrow $$ XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL = $${1 \over {{\omega _c}}}$$
$$ \Rightarrow $$ 75 = $${1 \over {\left( {2\pi F} \right)C}}$$
$$ \Rightarrow $$ C = $${1 \over {\left( {2\pi \times 50} \right)75}}$$
Also given, C = $$\left( {{n \over {3\pi }}} \right)$$ $$\mu $$F
$$ \therefore $$ $${1 \over {\left( {2\pi \times 50} \right)75}} = {{n \times {{10}^{ - 6}}} \over {3\pi }}$$
$$ \Rightarrow $$ n = 400
cos $$\phi $$ = 0.8 = $${R \over {\sqrt {{R^2} + X_L^2} }}$$ = $${R \over Z}$$
We know,
Power, P = $${{V_{rms}^2} \over {{Z^2}}} \times R$$
$$ \Rightarrow $$ 400 = $${{{{\left( {250} \right)}^2} \times 0.8Z} \over {{Z^2}}}$$
$$ \Rightarrow $$ Z = 125
$$ \therefore $$ R = 0.8 $$ \times $$ 125 = 100 $$\Omega $$
As Z2 = $$X_L^2 + {R^2}$$
$$ \Rightarrow $$ $${\left( {125} \right)^2} = X_L^2 + {\left( {100} \right)^2}$$
$$ \Rightarrow $$ XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL = $${1 \over {{\omega _c}}}$$
$$ \Rightarrow $$ 75 = $${1 \over {\left( {2\pi F} \right)C}}$$
$$ \Rightarrow $$ C = $${1 \over {\left( {2\pi \times 50} \right)75}}$$
Also given, C = $$\left( {{n \over {3\pi }}} \right)$$ $$\mu $$F
$$ \therefore $$ $${1 \over {\left( {2\pi \times 50} \right)75}} = {{n \times {{10}^{ - 6}}} \over {3\pi }}$$
$$ \Rightarrow $$ n = 400
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