Question

When $V = 100\sin \omega t$ is applied across a series (R−L−C) circuit, at resonance the current in resistance ($R = 100\Omega $) is $i = {i_0}\sin \omega t$, then power dissipation in circuit is:-
A) $50W$
B) $100W$
C) $25W$
D) Can′t be calculated

Answer 1

Hint: Firstly we will calculate root mean square value of voltage. After calculating it, we will find power dissipated using a suitable formula. Therefore, you have to remember both the formulas used here.

Formula Used:
As we know, Formula for calculating voltage is given by
${V_{RM{S_{}}}} = \dfrac{{{V_0}}}{{\sqrt 2 }}$
Here, ${V_{RMS}}$ is the root mean square voltage and ${V_0}$ is the applied voltage.
Also, formula used for calculating power dissipation is given by
$P = \dfrac{{{V^{^2}}}}{R}$
Here, $P$ is the power dissipated, $V$ is the voltage which is also known as root mean square voltage and $R$ is the resistance on the current.

Complete step by step solution:
Now, We are given that
${V_0} = 100\sin \omega t$ (Voltage applied)
$R = 100\Omega $ (Resistance on current)
As we know, ${V_{RMS}}$ = $\dfrac{{{V_0}}}{{\sqrt 2 }}$
$\therefore $ ${V_{RMS}} = \dfrac{{100}}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{{100}}{{\sqrt 2 }}\dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
$ \Rightarrow \dfrac{{100\sqrt 2 }}{2}$
$ \Rightarrow 50\sqrt 2 $
Now, power dissipated is given by
$P = \dfrac{{{V^2}}}{R}$
$ \Rightarrow \dfrac{{{{(50\sqrt 2 )}^2}}}{{50}}$
$ \Rightarrow \dfrac{{2500\times 2}}{{100}}$
$ \Rightarrow 25\times2$
$ \Rightarrow 50W$

Therefore, (A) is the correct option.

Additional Information:
Now, let us talk about ${V_{RMS}}$ .
${V_{RMS}}$ Is defined as the square root of the mean square of instantaneous values of voltage. Here, $V$ stands for the voltage and $RMS$ stands for root-mean Square. We use the term ${V_{RMS}}$ for the sinusoidal waveform. The term ${V_{RMS}}$ is used to describe the peak of voltage in current.

Also, the term $RMS$ is used to express the average voltage in an AC system.
Now, let us know the application of root mean square to voltage. For this,
Let, $V = {V_0}\sin \omega t$
Here, $V$ is the voltage, ${V_0}$ is the peak voltage, $f$ is the frequency and t is the time taken.
Now, for rms value of voltage we are given that,
${V_{RMS}} = \sqrt {\dfrac{1}{{{T_2} - {T_1}}}\int\limits_{{T_1}}^{{T_2}} {[{V_0}} \sin \omega t{]^2}dt} $
Now, we know that ${V_0}$ is constant here
$\therefore $ ${V_{RMS}} = \sqrt {\dfrac{1}{{{T_2} - {T_1}}}\int\limits_{{T_1}}^{{T_2}} {{{[\sin \omega t]}^2}dt} } $
$ \Rightarrow {V_0}\sqrt {\dfrac{1}{{{T_2} - {T_1}}}\int\limits_{{T_1}}^{{T_2}} {\dfrac{{1 - \cos (2\omega t)}}{2}dt} } $
Now we will integrate the terms in the square root,
${V_{RMS}} = {V_0}\sqrt {\dfrac{1}{{{T_2} - {T_1}}}[\dfrac{t}{2} - \dfrac{{\sin (2\omega t)}}{{4\omega }}]_{{T_1}}^{{T_2}}} $
As we know, in the sinusoidal waveform, intervals are a whole number of complete cycles. Therefore, the term will cancel out and we get,
${V_{RMS}} = {V_0}\sqrt {\dfrac{1}{{{T_2} - {T_1}}}[\dfrac{t}{2}]_{{T_2}}^{{T_1}}} $
$ \Rightarrow {V_0}\sqrt {\dfrac{1}{{{T_2} - {T_1}}}[\dfrac{{{T_2} - {T_1}}}{2}]} $
$ \Rightarrow \dfrac{V_0}{\sqrt{2}}$
Which is the formula for ${V_{RMS}}$ .

Note: The RMS value of a sinusoidal waveform gives the same heating effect as a DC current of the same value gives. Thus, we can say, if a direct current I, passes through a resistance of R ohms, the DC power consumed by the resistor will be, therefore, ${I^2}$R watts.