91Ó°ÊÓ

The frequency is, $f=60.0 \text{Hz}$.

The maximum current is, $I=300 \text{mA}=0.3 \text{A}$

The maximum emf is, ${\mathrm{Î\mu }}_m=30.0 \text{V}$

$V_L=2×V_R$

$V_L=2×V_C$

The voltage across the inductor in terms of the voltage across the capacitor and resistor. Using the formula for phase constant, find the angle by which the current lags the generator emf. Find the resistance of the circuit by using Ohm’s law.

Formulae are as follows:

$\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{Ï•}\mathbf{=}\left(\frac{V_L-V_C}{V_R}\right)$

$\mathit{V}\mathbf{=}\mathit{I}\mathit{R}$

Where,

Vis the potential difference.

Iis the current.

Ris the resistance.

The phase constant is given as,

$\tan \mathrm{Ï•}=\left(\frac{V_L-V_C}{V_R}\right)$

But,$$\begin{gathered} V_L=2×V_R \\ {V}_R=\frac{V_L}{2} \end{gathered}$$

Also,$$\begin{gathered} V_L=2×V_C \\ {V}_C=\frac{V_L}{2} \end{gathered}$$

Therefore,

$\begin{array}{rcl}\tan \mathrm{Ï•}& =& \left(\frac{V_L-\left(\frac{V_L}{2}\right)}{\left(\frac{V_L}{2}\right)}\right)\\ \tan \mathrm{Ï•}& =& \left(\frac{\frac{V_L}{2}}{\frac{V_L}{2}}\right)\\ \tan \mathrm{Ï•}& =& 1.00\\ & & \end{array}$

$$\begin{gathered} \mathrm{Ï•}={\tan }^{-1}\left(1.00\right) \\ \mathrm{Ï•}=45.0^0 \end{gathered}$$

Hence, the angle by which the current lags the generator emf is $45°$

By Ohm’s law,

$V=IR$

But,

$V={\mathrm{Î\mu }}_m\cos \mathrm{Ï•}$

Thus,

${\mathrm{Î\mu }}_m\cos \mathrm{Ï•}=IR$

Rearranging the terms,

$R=\frac{{\mathrm{Î\mu }}_m\cos \mathrm{Ï•}}{I}$

Substitute all the value in the above equation,

$\begin{array}{rcl}R& =& \frac{30.0V×\cos 45°}{0.3A}\\ & =& 70.7\mathrm{Î\copyright }\end{array}$

Hence, the resistance of the circuit when generator emf is $30.0V$and current is 300mA is$70.7\mathrm{Î\copyright }$

By using the given relation between voltages across the capacitor, resistor, and inductor, and the formula for phase constant, found the angle between the voltage and current. Using Ohm’s law, found the required resistance.