Question 234: AC fundamentals

Question

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 $${1 \over x} 	imes {10^{ - 3}}$$ F. Then the value of x  is ____________.

Accepted Answer

Given,
Inductance, L = 30 mH
Resistance, R = 1 $$\Omega$$
Angular frequency, $$\omega$$ = 300 rad/s
We know that in L-C-R circuit, $$	an \phi  = {{{X_C} - {X_L}} \over R}$$
where, $$\phi$$ = phase angle = 45$$^\circ$$
X
C
= capacitive reactance = $${1 \over {\omega C}}$$
X
L
= inductive reactance = $$\omega$$L
$$\Rightarrow$$ $$	an 45^\circ  = {{{X_C} - {X_L}} \over R}$$
$$ \Rightarrow {X_C} - {X_L} = R$$ [$$\because$$ tan 45$$^\circ$$ = 1]
$$ \Rightarrow {1 \over {\omega C}} - \omega L = R \Rightarrow {1 \over {\omega C}} - 300 	imes 30 	imes {10^{ - 3}} = 1$$
$$ \Rightarrow {1 \over {\omega C}} = 10 \Rightarrow \omega C = {1 \over {10}}$$
$$ \Rightarrow C = {1 \over {10\omega }} \Rightarrow C = {1 \over {10 	imes 300}}$$
$$ \Rightarrow C = {1 \over 3} 	imes {10^{ - 3}}F$$ .... (i)
According to question, the value of capacitance is $${1 \over x} 	imes {10^{ - 3}}F$$. So, on comparing it with Eq. (i), we can say x = 3.

AC fundamentals

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 ____________.

Integer / Numeric Answer Type