Question 33 of 190

In the given circuit the $AC$ source has $\omega =100rad{s}^{-1}$. Considering the inductor and capacitor to be ideal, what will be the current $I$ flowing through the circuit?

A$5.9A$B$6A$C$0.94A$D$3.16A$

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Answer: (D) $3.16A$

${Z}_{C}=\sqrt{{\left(\dfrac{1}{\omega C}\right)}^{2}+{R}^{2}}$
$=\sqrt{{\left(\dfrac{1}{100\times 100\times {10}^{-6}}\right)}^{2}+{100}^{2}}$
$=\sqrt{{100}^{2}+{100}^{2}}$
$=100\sqrt{2}\Omega$
${Z}_{L}=\sqrt{(\omega L{)}^{2}+{R}^{2}}=\sqrt{(100\times 0.5{)}^{2}+{50}^{2}}=50\sqrt{2}\Omega$
${i}_{L}=\dfrac{200}{{Z}_{L}}=\dfrac{200}{50\sqrt{2}}=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\text{A}$
${i}_{C}=\dfrac{200}{{Z}_{C}}=\dfrac{200}{100\sqrt{2}}=\sqrt{2}\text{A}$
$\cos {\phi }_{1}=\dfrac{100}{100\sqrt{2}}=\dfrac{1}{\sqrt{2}},{\phi }_{1}=45^{\circ}$
$\cos {\phi }_{2}=\dfrac{50}{50\sqrt{2}}=\dfrac{1}{\sqrt{2}},{\phi }_{2}=45^{\circ}$

$I=\sqrt{{I}_{C}^{2}+{I}_{L}^{2}};I=\sqrt{8+2};I=\sqrt{10}=3.16A$
This question is bonus as the correct answer was not available for students to mark as option in original paper by NTA.

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