KCET · Physics · Alternating Current
In the given circuit the peak voltage across \(C\), \(L\) and \(R\) are \(30 \mathrm{~V}, 110 \mathrm{~V}\) and \(60 \mathrm{~V}\), respectively. The rms value of the applied voltage is
- A \(100 \mathrm{~V}\)
- B \(200 \mathrm{~V}\)
- C \(70.7 \mathrm{~V}\)
- D \(141 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(70.7 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, \(V_{C}=30 \mathrm{~V}, V_{L}=110 \mathrm{~V}\) and \(V_{R}=60 \mathrm{~V}\) The peak voltage across series \(L-C-R\) circuit is given by
\(\begin{aligned} V_{O} &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ &=\sqrt{(60)^{2}+(110-30)^{2}} \\ &=\sqrt{(60)^{2}+(80)^{2}}=100 \mathrm{~V} \end{aligned}\)
\(\therefore\) Rms value of voltage, \(V_{\text {rms }}=\)
\(\begin{aligned} V_{O} &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ &=\sqrt{(60)^{2}+(110-30)^{2}} \\ &=\sqrt{(60)^{2}+(80)^{2}}=100 \mathrm{~V} \end{aligned}\)
\(\therefore\) Rms value of voltage, \(V_{\text {rms }}=\)
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