Question

In the adjoining circuit diagram of a $LCR$ , find the heat developed in $80s$ and the amplitude of wattless current.

• A. $4000J,3A$
• B. $8000J,3A$
• C. $4000J,4A$
• D. $8000J,5A$

Answer 1

Answer: (A)

For any series $LCR$ circuit, the heat generated in the circuit is given by :
$H=P\times t$
Where $P$ is the power generated by the circuit and $t$ denotes the period of time for which the current flows in the circuit.
Also, $P=\left(\right.i_{r m s}\left(\left.\right)^{2}\cdot R$
Where, $R=4\Omega$
Hence, $H=\left(\right.i_{r m s}\left(\left.\right)^{2}\cdot R\times t$
CALCULATION OF CURRENT :
Maximum value of current in the given circuit is :
$i_{o}=\frac{V_{o}}{Z}$
where $V_{o}=25$ as given $E=25sin\left(100 \pi t + \left(\pi \right)/2\right)$ and $Z$ is given by:
$\Rightarrow Z=\sqrt{\left(R\right)^{2} + \left(X_{L}\right)^{2} - \left(X_{C}\right)^{2}\right)}$
$\Rightarrow Z=\sqrt{\left(4\right)^{2} + \left(7 - 4\right)^{2}}$
$\Rightarrow Z=\sqrt{16 + 9}$
$\Rightarrow Z=5\Omega$
Now, we need to calculate RMS value as: $i_{rms}=\frac{i_{0}}{\sqrt{2}}=\frac{\left(\right. V_{0} / Z \left.\right)}{\sqrt{2}}=\frac{25 / 5}{\sqrt{2}}A$
CALCULATION OF HEAT:
Putting these value in the above heat equation, we get
$\Rightarrow H=\left(\right.i_{rms}\left(\left.\right)^{2}\cdot Rt$
$\Rightarrow H=\left(\frac{25}{5 \sqrt{2}}\right)^{2}\times 4\times 80=4000J$

CALCULATION OF WATTLESS CURRENT:
The Amplitude of wattless current is given as:
$I_{WL}=I_{0}sin\phi$
Where, $\phi=\left(tan\right)^{- 1}\left(\right.\frac{X_{L} - X_{c}}{R}\left.\right)$
Substituting the values from question:
$\Rightarrow \phi=\left(tan\right)^{- 1}\left(\right.\frac{7 - 4}{4}\left.\right)$
$\Rightarrow \phi=37^{o}$
$\Rightarrow I_{WL}=I_{0}sinϕ$
$\Rightarrow I_{WL}=\frac{V_{0}}{Z}sin\left(\phi\right)$
$\Rightarrow I_{WL}=\frac{25}{5}\times sin37^\circ =3A$