KCET · Physics · Current Electricity
What will be the reading in the voltmeter and ammeter of the circuit shown?
- A \(90 \mathrm{~V}, 2 \mathrm{~A}\)
- B \(0,2 \mathrm{~A}\)
- C \(90 \mathrm{~V}, 1 \mathrm{~A}\)
- D \(0,1 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(0,2 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given, \(X_{L}=X_{C}=4 \Omega\)
\(\therefore\) Voltage across \(X_{C}\) and \(X_{L}\) is same in magnitude but opposite in direction. So, net voltage is
\(V_{\text {net }}=V_{C}-V_{L}=0\)
Since, the voltmeter is connected across the series connection of capacitor and inductor. So, the reading of voltmeter is equal to \(V_{\text {net }}\) i.e., zero.
Current in the circuit,
\(\begin{aligned}I &=\frac{V}{Z} \\&=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \\&=\frac{90}{\sqrt{(45)^{5}+(4-4)^{2}}}=\frac{90}{45}=2 \mathrm{~A}\end{aligned}\)
\(\therefore\) Voltage across \(X_{C}\) and \(X_{L}\) is same in magnitude but opposite in direction. So, net voltage is
\(V_{\text {net }}=V_{C}-V_{L}=0\)
Since, the voltmeter is connected across the series connection of capacitor and inductor. So, the reading of voltmeter is equal to \(V_{\text {net }}\) i.e., zero.
Current in the circuit,
\(\begin{aligned}I &=\frac{V}{Z} \\&=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \\&=\frac{90}{\sqrt{(45)^{5}+(4-4)^{2}}}=\frac{90}{45}=2 \mathrm{~A}\end{aligned}\)
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