Question

In LCR series circuit, an alternating emf $e$ and current $i$ are given by the equations $e = 100\sin \left( {100t} \right)\,volt$, $i = 100\sin \left( {100 + \dfrac{\pi }{3}} \right)\,mA$. The average power dissipated in the circuit will be:
(A) $100\,W$
(B) $10\,W$
(C) $5\, W$
(D) $2.5\,W$

Answer 1

The average power dissipated in the LCR circuit will be determined by using the two formulas, sinusoidal voltage formula, and the sinusoidal current formula. By comparing the equation with the given equation in the question, the power dissipated can be determined.

Formula used:
Sinusoidal voltage is given by,
$e = {e_0}\sin \omega t$
Where, $e$ is the emf, ${e_0}$ is the peak emf, $\omega$ is the frequency and $t$ is the time.
Sinusoidal current is given by,
$i = {i_0}\sin \left( {\omega t + \phi } \right)$
Where, $i$ is the current, ${i_0}$ is the peak current, $\omega$ is the frequency, $t$ is the time and $\phi$ is the phase angle.
The rms value of the emf is,
${e_{rms}} = \dfrac{{{e_0}}}{{\sqrt 2 }}$
Where, ${e_{rms}}$ is the rms value of the emf and ${e_0}$ is the peak emf.
The rms value of the current is,
${i_{rms}} = \dfrac{{{i_0}}}{{\sqrt 2 }}$
Where, ${i_{rms}}$ is the rms value of the current and ${i_0}$ is the peak current.
The average power is given by,
$P = {e_{rms}} \times {i_{rms}} \times \cos \phi$
Where, $P$ is the power, ${e_{rms}}$ is the rms value of the emf, ${i_{rms}}$ is the rms value of the current and $\phi$ is the phase angle.

Complete step by step answer:
Given that,
The equation of the emf is, $e = 100\sin \left( {100t} \right)\,volt$.
The equation of the current is, $i = 100\sin \left( {100 + \dfrac{\pi }{3}} \right)\,mA$
Now,
Sinusoidal voltage is given by,
$e = {e_0}\sin \omega t\,................\left( 1 \right)$
Sinusoidal current is given by,
$i = {i_0}\sin \left( {\omega t + \phi } \right)\,.....................\left( 2 \right)$
By comparing the equation (1) with the given emf equation in the question,
$\Rightarrow {e_0}\sin \omega t = 100\sin \left( {100t} \right)$
Then the peak emf is,
${e_0} = 100\,V$
By comparing the equation (2) with the given current equation in the question,
$\Rightarrow {i_0}\sin \left( {\omega t + \phi } \right) = 100\sin \left( {100 + \dfrac{\pi }{3}} \right)\,mA$
Then the peak current is,
${i_0} = 100\,mA$
The phase angle is $\phi = \dfrac{\pi }{3}$
Now, the rms value of the emf is,
$\Rightarrow {e_{rms}} = \dfrac{{{e_0}}}{{\sqrt 2 }}$
By substituting the peak emf value in the above equation, then the above equation is written as,
$\Rightarrow {e_{rms}} = \dfrac{{100}}{{\sqrt 2 }}\,V$
Now, the rms value of the current is,
$\Rightarrow {i_{rms}} = \dfrac{{{i_0}}}{{\sqrt 2 }}$
By substituting the peak current value in the above equation, then the above equation is written as,
$\Rightarrow {i_{rms}} = \dfrac{{100}}{{\sqrt 2 }} \times {10^{ - 3}}\,A$
Now, the average power is given by,
$\Rightarrow P = {e_{rms}} \times {i_{rms}} \times \cos \phi$
By substituting the ${e_{rms}}$, ${i_{rms}}$ and $\phi$ values in the above equation, then
$\Rightarrow P = \dfrac{{100}}{{\sqrt 2 }} \times \dfrac{{100}}{{\sqrt 2 }} \times {10^{ - 3}} \times \cos \dfrac{\pi }{3}$
From trigonometry the value of $\cos \dfrac{\pi }{3}$ is $\dfrac{1}{2}$, then
$\Rightarrow P = \dfrac{{100}}{{\sqrt 2 }} \times \dfrac{{100}}{{\sqrt 2 }} \times {10^{ - 3}} \times \dfrac{1}{2}$
By solving the above equation then,
$\Rightarrow P = 2500 \times {10^{ - 3}}$
On simplification,
$\Rightarrow P = 2.5\,W$

$\therefore$ The average power dissipated in the circuit will be 2.5W. Hence, the option (D) is the correct answer.

Note:
An LCR circuit, also known as a resonant circuit, tuned circuit, or an RLC circuit, is an electrical circuit consisting of an inductor, capacitor, and resistor connected in series or parallel. But the resistance, current, and voltage phasors are always in phase.