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Physics Question on Alternating current

A resistor of resistance RR, capacitor of capacitance CC and inductor of inductance LL are connected in parallel to ACA C power source of voltage ϵ0sin⁡ωt\epsilon_{0} \sin \omega_{t}. The maximum current through the resistance is half of the maximum current through the power source. Then value of RR is

A

3∣ωC−1ωL∣\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}

B

3∣1ωC−ωL∣\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right|

C

5∣1ωC−ωL∣\sqrt{5}\left|\frac{1}{\omega C}-\omega L\right|

D

None of these

Answer

3∣ωC−1ωL∣\frac{\sqrt{3}}{\left|\omega C-\frac{1}{\omega L}\right|}

Explanation

Solution

iR0(iR0)2+(ic0−iL0)2=12\frac{i_{R_{0}}}{\sqrt{\left(i_{R_{0}}\right)^{2}+\left(i_{c_{0}}-i_{L_{0}}\right)^{2}}}=\frac{1}{2} ⇒ϵ0/R(ϵ0/R)2(ϵ0ωC−ϵ0ωL)2=12\Rightarrow \frac{\epsilon_{0} /R}{\sqrt{\left(\epsilon_{0} /R\right)^{2}\left(\epsilon_{0} \omega C-\frac{\epsilon_{0}}{\omega_{L}}\right)^{2}}}=\frac{1}{2} ⇒R=3(ωC−1ωL)\Rightarrow R=\frac{\sqrt{3}}{\left(\omega C-\frac{1}{\omega L}\right)}