Question

An inductor of \(\frac{5}{\pi}\,H\), a capacitor of \(\frac{50}{\pi}\,\mu F\) and a resistor of \(400\,\Omega\) are connected in series across an ac voltage \[ v = 140 \sin(100\pi t)\,V. \] Calculate:
• (I) impedance of the circuit
• (II) rms value of current in the circuit} (Take \(\sqrt{2}=1.4\))

Answer 1

Given: \(\sqrt{2}=1.4\)

Given

\[ L=\frac{5}{\pi}\;H,\qquad C=\frac{50}{\pi}\times10^{-6}\;F,\qquad R=400\;\Omega \]

From \[ v=140\sin(100\pi t) \]

Peak voltage, \[ V_0=140\;V \]

Angular frequency, \[ \omega=100\pi\;rad\,s^{-1} \]

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Step 1: Calculate Inductive Reactance

\[ X_L=\omega L \]

\[ X_L=(100\pi)\times\frac{5}{\pi}=500\;\Omega \]

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Step 2: Calculate Capacitive Reactance

\[ X_C=\frac{1}{\omega C} \]

\[ X_C=\frac{1}{100\pi\times\frac{50}{\pi}\times10^{-6}} =\frac{1}{5\times10^{-3}} =200\;\Omega \]

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Step 3: Calculate Impedance

Net reactance,

\[ X=X_L-X_C=500-200=300\;\Omega \]

Impedance,

\[ Z=\sqrt{R^2+X^2} \]

\[ Z=\sqrt{400^2+300^2} =\sqrt{160000+90000} =\sqrt{250000} =500\;\Omega \]

Hence,

\[ \boxed{Z=500\;\Omega} \]

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Step 4: Calculate RMS Current

RMS voltage,

\[ V_{\rm rms}=\frac{V_0}{\sqrt2} =\frac{140}{1.4}=100\;V \]

Using,

\[ I_{\rm rms}=\frac{V_{\rm rms}}{Z} \]

\[ I_{\rm rms}=\frac{100}{500}=0.2\;A \]

Hence,

\[ \boxed{I_{\rm rms}=0.2\;A} \]

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Final Answer

• (i) Impedance: \[ \boxed{500\;\Omega} \]
• (ii) RMS current: \[ \boxed{0.2\;A} \]