Electromagnetic Induction and Alternating Current 6 Question 11
11. In the given circuit, the $AC$ source has $\omega=100 rad / s$. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)
(2012)
(a) the current through the circuit, $I$ is $0.3 A$
(b) the current through the circuit, $I$ is $0.3 \sqrt{2} A$
(c) the voltage across $100 \Omega$ resistor $=10 \sqrt{2} V$
(d) the voltage across $50 \Omega$ resistor $=10 V$
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Answer:
Correct Answer: 11. $(a, c)$
Solution:
Circuit 1
$$ \begin{aligned} X _C & =\frac{1}{\omega C}=100 \Omega \\ \therefore \quad Z _1 & =\sqrt{(100)^{2}+(100)^{2}}=100 \sqrt{2} \Omega \\ \varphi _1 & =\cos ^{-1} \frac{R _1}{Z _1}=45^{\circ} \end{aligned} $$
In this circuit, current leads the voltage.
$$ \begin{aligned} I _1 & =\frac{V}{Z _1}=\frac{20}{100 \sqrt{2}}=\frac{1}{5 \sqrt{2}} A \\ V _{100 \Omega} & =(100) I _1=(100) \frac{1}{5 \sqrt{2}} V=10 \sqrt{2} V \end{aligned} $$
Circuit 2
$$ \begin{aligned} X _L & =\omega L=(100)(0.5)=50 \Omega \\ Z _2 & =\sqrt{(50)^{2}+(50)^{2}}=50 \sqrt{2} \Omega \\ \varphi _2 & =\cos ^{-1} \frac{R _2}{Z _2}=45^{\circ} \end{aligned} $$
In this circuit, voltage leads the current.
$$ \begin{aligned} I _2 & =\frac{V}{Z _2}=\frac{20}{50 \sqrt{2}}=\frac{\sqrt{2}}{5} A \\ V _{50 \Omega} & =(50) I _2=50 \quad \frac{\sqrt{2}}{5}=10 \sqrt{2} V \end{aligned} $$
Further, $I _1$ and $I _2$ have a mutual phase difference of $90^{\circ}$.
$$ \therefore \quad I=\sqrt{I _1^{2}+I _2^{2}}=0.34 $$
