Question

An inductor and a resistor are connected in series to an $AC$ source. The current in circuit is $500\, mA$, if the applied $AC$ voltage is $8 \sqrt{2} V$ at a frequency of $\frac{175}{\pi} Hz$ and the current in the circuit is $400\, mA$, if the same $AC$ voltage at a frequency of $\frac{225}{\pi} Hz$ is applied. The values of the inductance and the resistance are respectively

• A. $60\, mH,\, 71\, \Omega$
• B. $\sqrt{60}\, mH,\, 71\, \Omega$
• C. $\sqrt{60}\, mH,\, \sqrt{71}\, \Omega$
• D. $60\, mH,\, \sqrt{71}\, \Omega$

Answer 1

The Correct Option is D

For an $L-R \text { circuit, } I=\frac{V}{Z}=\frac{V}{\sqrt{R^{2}+L^{2} \omega^{2}}}$
$\Rightarrow R^{2}+L^{2} \omega^{2} =\left(\frac{V}{I}\right)^{2}$
Here, $I_{1} =500 \times 10^{-3} A$
$\omega_{1}=\frac{175}{\pi} \times 2 \pi \frac{ rad }{ s }=350 \frac{ rad }{ s }$
$V_{1} =8 \sqrt{2}$
$\Rightarrow R^{2}+L^{2}(350)^{2}=\left(\frac{8 \sqrt{2}}{500 \times 10^{-3}}\right)^{2}$
$\Rightarrow R^{2}+L^{2}(350)^{2}=512$ ...(i)
and $R^{2}+L^{2}(550)^{2} =800$ ...(ii)
Solving, we get $R=\sqrt{71} \,\Omega \text { and } L=60\, mH$