Question: Q. 8. A $200 \mathrm{mH}$ (pure) inductor and a $5 \mu \mathrm{F}$ (pure) capacitor are connected one by one, across a sinusoidal $a c$ voltage source of $V=[70.7 \sin (1000$ $t)]$ voltage. Obtain the expression for the current in each case.
A [Foreign, 2016]
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Solution:
Ans. For the applied voltage
$$ \begin{aligned} V & =70.7 \sin (1000 t), \text { we have } \ V_{0} & =70.7 \text { volts } \end{aligned} $$
For the inductor
$$ \begin{aligned} i_{o} & =\frac{V_{0}}{\omega L}=\frac{70.7}{1000 \times 200 \times 10^{-3}} \mathrm{~A} \ & =35.35 \times 10^{-2} \mathrm{~A} \ & =0.3535 \mathrm{~A} \end{aligned} $$
$\therefore$ Expression for current is
$$ \begin{equation*} i=(0.3535) \sin \left(1000 t-\frac{\pi}{2}\right) \tag{Q.} \end{equation*} $$
For the capacitor
$$ \begin{aligned} i_{0} & =\frac{V_{0}}{\left(\frac{1}{\omega C}\right)}=V_{0} \cdot \omega C \ & =70.7 \times 1000 \times 5 \times 10^{-6} \mathrm{~A} \ & =353.5 \times 10^{-3} \mathrm{~A}=0.3535 \mathrm{~A} \end{aligned} $$
$\therefore$ Expression for current is
$$ I=0.3535 \sin \left(1000 t+\frac{\pi}{2}\right) $$
[CBSE Marking Scheme, 2016]
[AI Q. 9. A circuit containing an $80 \mathrm{mH}$ inductor and a $250 \mu \mathrm{F}$ capacitor in series connected to a $240 \mathrm{~V}$, $100 \mathrm{rad} / \mathrm{s}$ supply. The resistance of the circuit is negligible.
(i) Obtain $r m s$ value of current.
(ii) What is the total average power consumed by the circuit?
A [Delhi I, II, III 2015]
Ans. (i)
$$ \begin{aligned} X_{L} & =\omega L=100 \times 80 \times 10^{-3}= \ X_{C} & =\frac{1}{\omega C}=\frac{1}{100 \times 250 \times 10^{-6}} \ & =40 \Omega \end{aligned} $$
Total Impedance $(Z)=X_{C}-X_{\mathrm{L}}$
$$ =32 \Omega $$
$$ I_{r m s}=\frac{240}{32} \mathrm{~A} \Leftrightarrow 7.5 \mathrm{~A} $$
