AP EAMCET · PHYSICS · Alternating Current
An inductor and a resistor are connected in series to an AC source. The current in circuit is \(500 \mathrm{~mA}\), if the applied AC voltage is \(8 \sqrt{2} \mathrm{~V}\) at a frequency of \(\frac{175}{\pi} \mathrm{Hz}\) and the current in the circuit is \(400 \mathrm{~mA}\), if the same AC voltage at a frequency of \(\frac{225}{\pi} \mathrm{Hz}\) is applied. The values of the inductance and the resistance are respectively
- A \(60 \mathrm{mH}, 71 \Omega\)
- B \(\sqrt{60} \mathrm{mH}, 71 \Omega\)
- C \(\sqrt{60} \mathrm{mH}, \sqrt{71} \Omega\)
- D \(60 \mathrm{mH}, \sqrt{71} \Omega\)
Answer & Solution
Correct Answer
(D) \(60 \mathrm{mH}, \sqrt{71} \Omega\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { For an } L-R \text { circuit, } I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+L^2 \omega^2}} \\ & \Rightarrow \quad R^2+L^2 \omega^2=\left(\frac{V}{I}\right)^2 \\ & \text { Here, } \quad I_1=500 \times 10^{-3} \mathrm{~A} \\ & \omega_1=\frac{175}{\pi} \times 2 \pi…
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