Question

A series AC circuit containing an inductor (20mH), a capacitor (120 $\mu H$ ), and a resistor (60 $\Omega$ ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is:
(1) $5\cdot 65\times {{10}^{2}}J$
(2) $5\cdot 17\times 10{}^{2}J$
(3) $2\cdot 26\times {{10}^{3}}J$
(4) $3\cdot 39\times {{10}^{3}}J$

Answer 1

An LCR circuit is an electrical circuit consisting of a resistor, inductor, Capacitor, connected in series or in parallel. To find out power dissipation in circuit, first of all we have to calculate total impedance using $Z=\sqrt{{{(R)}^{2}}+{{({{X}_{C}}-{{X}_{L}})}^{2}}}$ and after that power factor,
$\Rightarrow \cos \phi =\dfrac{R}{Z}$ . Putting these values in $P={{E}_{v}}{{I}_{v}}\cos \phi .t$ this relation to find out power dissipated in the circuit.

COmplete step by step solution
Here, $R=60\Omega$
$\Rightarrow L=20mH$
$\Rightarrow 1H=1000mH$ $\therefore$ $L=20\times {{10}^{-3}}H$
$\Rightarrow C=120\mu F$
$\Rightarrow 1F={{10}^{6}}\mu F$ , $\therefore C=120\times {{10}^{-6}}F$
$\Rightarrow \nu =50Hz,{{E}_{\nu }}=24V,{{I}_{\nu }}=?,{{P}_{{}}}=?$
Impedance in case of LCR circuit is given as:
$\begin{aligned} &\Rightarrow Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}} \\\ & where,X{}_{L}=\omega L \\\ &\Rightarrow {{X}_{C}}=\dfrac{1}{\omega C} \\\ \end{aligned}$ $\to (1)$
To find out the value of $\omega$ we will use $\omega =2\pi \nu$ relation. Now, put all the values in this relation.
$\begin{aligned} &\Rightarrow 2\times \pi \times 50 \\\ &\Rightarrow 100\pi \\\ \end{aligned}$
Now, use this value to find out ${{X}_{L}}and{{X}_{C}}$
$\begin{aligned} &\Rightarrow {{X}_{L}}=\omega L \\\ &\Rightarrow 100\pi \times 20\times {{10}^{-3}} \\\ &\Rightarrow 2\pi \Omega \\\ \end{aligned}$
And,
$\begin{aligned} &\Rightarrow {{X}_{C}}=\dfrac{1}{\omega C} \\\ &\Rightarrow \dfrac{1}{100\pi \times 120\times {{10}^{-6}}} \\\ &\Rightarrow \dfrac{{{10}^{6}}}{12000\pi } \\\ &\Rightarrow \dfrac{{{10}^{3}}\times 7}{12\times 22} \\\ &\Rightarrow \dfrac{7000}{264}=26.51\Omega \\\ \end{aligned}$
Now, find the difference between ${{X}_{C}}and{{X}_{L}}$
$\begin{aligned} &\Rightarrow {{X}_{C}}-{{X}_{L}}=26.51-2\pi =26.51-2\times \dfrac{22}{7} \\\ &\Rightarrow 26.51-6.28=20.23\approx 20 \\\ \end{aligned}$ $$ $\to (2)$
Put the value of R and equation (2) in equation (1)
We get, $\begin{aligned} &\Rightarrow \sqrt{{{(60)}^{2}}+{{(20)}^{2}}} \\\ &\Rightarrow \sqrt{3600+400}=\sqrt{4000}=20\sqrt{10}\Omega \\\ \end{aligned}$
$\Rightarrow Z=20\sqrt{10}\Omega$
Power factor is given by:
$\Rightarrow \cos \phi =\dfrac{R}{Z}$
$\Rightarrow \dfrac{60}{20\sqrt{10}}=\dfrac{3}{\sqrt{10}}$
$$ Average power dissipated/cycle in RLC circuit is given by: $P={{E}_{\nu }}I{}_{\nu }\cos \phi$ $\to (A)$
And, ${{I}_{\nu }}=\dfrac{V}{Z}$
$\Rightarrow \dfrac{24}{20\sqrt{10}}=\dfrac{6}{5\sqrt{10}}$
Now, put all the values in equation (A)
$\Rightarrow P=24\times$ $\dfrac{6}{5\sqrt{10}}\times \dfrac{3}{\sqrt{10}}$
$\Rightarrow \dfrac{24\times 6\times 3}{5\sqrt{10}\sqrt{10}}=\dfrac{432}{50}=8.64Watt$
$\begin{aligned} &\Rightarrow Q=P.t=8.64\times 60 \\\ &\Rightarrow 5.18\times {{10}^{2}}J \\\ \end{aligned}$
$\therefore$ Option (2) is correct.

Note
RLC circuits have many applications as oscillator circuits. It can be used as filters, tuned circuits etc. While doing numerical be careful with the S.I. units. Concept of impedance and voltage should be clear. For more information refer to chapter Alternating circuit.