MHT CET · Physics · Alternating Current
When 100 V d.c. is applied across a solenoid, a current of 1 A flows in it. When 100 a.c. is applied across it, the current drops to 0.5 A . If the frequency is 50 Hz , the impedance and inductance is
- A \(200 \Omega, \sqrt{3} / \pi \mathrm{H}\)
- B \(100 \Omega, \sqrt{3} \mathrm{H}\)
- C \(200 \Omega, 1 \mathrm{H}\)
- D \(100 \Omega, 1 \mathrm{H}\)
Answer & Solution
Correct Answer
(A) \(200 \Omega, \sqrt{3} / \pi \mathrm{H}\)
Step-by-step Solution
Detailed explanation
When dc is applied
\(i=\frac{V}{R} \Rightarrow 1=\frac{100}{R} \Rightarrow R=100 \Omega\)
When ac is applied
\(i=\frac{V}{Z} \Rightarrow 0.5=\frac{100}{Z} \Rightarrow Z=200 \Omega\)
Hence \(Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+4 \pi^2 v^2 L^2}\)
\(\begin{aligned}
& \Rightarrow(200)^2=(100)^2+4 \pi^2(50)^2 L^2 \\
& \Rightarrow L=0.55 H
\end{aligned}\)
\(i=\frac{V}{R} \Rightarrow 1=\frac{100}{R} \Rightarrow R=100 \Omega\)
When ac is applied
\(i=\frac{V}{Z} \Rightarrow 0.5=\frac{100}{Z} \Rightarrow Z=200 \Omega\)
Hence \(Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+4 \pi^2 v^2 L^2}\)
\(\begin{aligned}
& \Rightarrow(200)^2=(100)^2+4 \pi^2(50)^2 L^2 \\
& \Rightarrow L=0.55 H
\end{aligned}\)
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