A series RLC circuit is driven in such a way that the maximum voltage across the inductor is $1.50times$the maximum voltage across the capacitor andlocalid="1664189666110" $\mathbf{2}\mathbf{}\mathit{t}\mathit{i}\mathit{m}\mathit{e}\mathit{s}$the maximum voltage across the resistor. (a) What is $\varphi$for the circuit? (b) Is the circuit inductive, capacitive, or in resonance? The resistance islocalid="1664189661326" $\mathbf{49}\mathbf{.}\mathbf{9}\mathbf{}\mathit{o}\mathit{h}\mathit{m}$, and the current amplitude islocalid="1664189655682" $\mathbf{200}\mathit{m}\mathit{A}$. (c) What is the amplitude of the driving emf?

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-

$\text{tan}\varphi =\frac{V_L-V_c}{V_R}$ (ii)

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

.$\text{tan}\varphi =\frac{V_L-V_c}{V_R}$

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

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

$\text{tan}\varphi =\frac{V_L-\frac{V_L}{\text{1.5}}}{\frac{V_L}{\text{2}}}$

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

Hence,thephase angle is $0.588rad$.

Since,$\varphi >0$the circuit should be inductive.

Hence, the circuit is inductive.

The voltage across the resistance is as follows:

$$\begin{gathered} V_R=IR \\ {V}_R=\left(0.2A\right)\times \left(49.9\Omega \right) \\ {V}_R=10V \end{gathered}$$

Now,

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

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$12V$.