In an LCR series circuit, an inductor 30 mH and a resistor 1 $\Omega$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45$^\circ$ is ${1 \over x} \times {10^{ - 3}}$ F. Then the value of x is ____________.
Answer (integer)
3
Solution
Given,<br/><br/>Inductance, L = 30 mH<br/><br/>Resistance, R = 1 $\Omega$<br/><br/>Angular frequency, $\omega$ = 300 rad/s<br/><br/>We know that in L-C-R circuit, $\tan \phi = {{{X_C} - {X_L}} \over R}$<br/><br/>where, $\phi$ = phase angle = 45$^\circ$<br/><br/>X<sub>C</sub> = capacitive reactance = ${1 \over {\omega C}}$<br/><br/>X<sub>L</sub> = inductive reactance = $\omega$L<br/><br/>$\Rightarrow$ $\tan 45^\circ = {{{X_C} - {X_L}} \over R}$<br/><br/>$\Rightarrow {X_C} - {X_L} = R$ [$\because$ tan 45$^\circ$ = 1]<br/><br/>$$ \Rightarrow {1 \over {\omega C}} - \omega L = R \Rightarrow {1 \over {\omega C}} - 300 \times 30 \times {10^{ - 3}} = 1$$<br/><br/>$\Rightarrow {1 \over {\omega C}} = 10 \Rightarrow \omega C = {1 \over {10}}$<br/><br/>$\Rightarrow C = {1 \over {10\omega }} \Rightarrow C = {1 \over {10 \times 300}}$<br/><br/>$\Rightarrow C = {1 \over 3} \times {10^{ - 3}}F$ .... (i)<br/><br/>According to question, the value of capacitance is ${1 \over x} \times {10^{ - 3}}F$. So, on comparing it with Eq. (i), we can say x = 3.
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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