MHT CET · Physics · Alternating Current
An a.c. voltage source \(\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}\). is connected across resistance ' R ' and capacitance ' \(C\) ' in series. It is given that \(R=\frac{1}{\omega \mathrm{c}}\) and the peak current is \(\mathrm{I}_0\). If the angular frequency of the voltage source is changed to \(\left(\frac{\omega}{\sqrt{3}}\right)\), then the new peak current in the circuit is
- A \(\frac{\mathrm{I}_0}{2}\)
- B \(\frac{\mathrm{I}_0}{\sqrt{2}}\)
- C \(\sqrt{2} \mathrm{I}_0\)
- D \(\sqrt{3} \mathrm{I}_0\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{I}_0}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Given: \(R=\frac{1}{\omega \mathrm{C}}=\mathrm{X}_{\mathrm{c}}\).
\(\begin{aligned}
\therefore \quad Z & =\sqrt{R^2+X_C^2}=\sqrt{2} R \\
I_0 & =\frac{V_0}{Z}=\frac{V_0}{\sqrt{2} R} \\
& \Rightarrow \frac{I_0}{\sqrt{2}}=\frac{V_0}{2 R}...(i)
\end{aligned}\)
When \(\omega\) becomes \(\frac{1}{\sqrt{3}}\) times, \(X_C\) will become \(\sqrt{3}\) times, i.e., \(\sqrt{3} R\).
\(\begin{array}{ll}
\therefore & Z^{\prime}=\sqrt{R^2+(\sqrt{3} R)^2}=2 R \\
\therefore & I_0^{\prime}=\frac{V_0}{Z^{\prime}}=\frac{V_0}{2 R}=\frac{I_0}{\sqrt{2}}...[From(i)]
\end{array}\)
\(\begin{aligned}
\therefore \quad Z & =\sqrt{R^2+X_C^2}=\sqrt{2} R \\
I_0 & =\frac{V_0}{Z}=\frac{V_0}{\sqrt{2} R} \\
& \Rightarrow \frac{I_0}{\sqrt{2}}=\frac{V_0}{2 R}...(i)
\end{aligned}\)
When \(\omega\) becomes \(\frac{1}{\sqrt{3}}\) times, \(X_C\) will become \(\sqrt{3}\) times, i.e., \(\sqrt{3} R\).
\(\begin{array}{ll}
\therefore & Z^{\prime}=\sqrt{R^2+(\sqrt{3} R)^2}=2 R \\
\therefore & I_0^{\prime}=\frac{V_0}{Z^{\prime}}=\frac{V_0}{2 R}=\frac{I_0}{\sqrt{2}}...[From(i)]
\end{array}\)
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