91Ó°ÊÓ

1. The maximum voltage across the inductor is $1.50times$the maximum voltage across the capacitor.
2. The maximum voltage across the inductor isrole="math" localid="1663219519545" $2times$ themaximum voltage across the resistor
3. Resistance is$49.9ohm$.
4. current is $200mA$.

Here, use the formula for phase angle in terms of voltage across inductance, the voltage across resistance, and resistance. And from the phase angle, decide whether the circuit is inductive or capacitive. Then, use the formula for amplitude voltage.

The electromotive force is given as-

$n=\sqrt{V_R^2+{\left(V_L-V_c\right)}^2}$ (i)

Here, n is EMF amplitude voltage, $V_R$is voltage across resistance, $V_L$is voltage across inductor, ${\mathit{V}}_{\mathbf{C}}$is voltage across capacitor.

Phase angle is given as-

$\text{tan}\mathrm{Ï•}=\frac{V_L-V_c}{V_R}$ (ii)

The phase angle of the LCR circuit is given as follows:

.$\text{tan}\mathrm{Ï•}=\frac{V_L-V_c}{V_R}$

Here$V_L$is the voltage across the inductor and$V_C$voltageacross the capacitor.

And$V_R=\frac{V_L}{\text{2}}$

$\text{tan}\mathrm{Ï•}=\frac{V_L-\frac{V_L}{\text{1.5}}}{\frac{V_L}{\text{2}}}$

$$\begin{gathered} \mathrm{ϕ}=33.7° \\ =0.588rad \end{gathered}$$

Hence,thephase angle is $0.588rad$.

Since,$\mathrm{Ï•}>0$the circuit should be inductive.

Hence, the circuit is inductive.

The voltage across the resistance is as follows:

$$\begin{gathered} V_R=IR \\ {V}_R=\left(0.2A\right)×\left(49.9\mathrm{Î\copyright }\right) \\ {V}_R=10V \end{gathered}$$

Now,

$$\begin{gathered} V_L=2V_R \\ {V}_L=2×10V \\ {V}_L=20V \end{gathered}$$

Now,

$$\begin{gathered} V_c=\frac{V_L}{1.5} \\ {V}_c=\frac{20V}{1.5} \\ {V}_c=13.5V \end{gathered}$$

Now amplitude is,

$$\begin{gathered} n=\sqrt{V_R^2+{\left(V_L-V_c\right)}^2} \\ n=\sqrt{{\left(9.98V\right)}^2+{\left(20V-13.3V\right)}^2} \\ n=12.0V \end{gathered}$$

Hence,the amplitude of driving emf is$12V$.