Answer:
The rms voltage across the inductor is 5.05 x 10⁻⁶ V.
Explanation:
Given;
inductance of the inductor, L = 85.2 mH = 85.2 x 10⁻³ H
capacitance of the capacitor, C = 5.27 pF = 5.27 x 10⁻¹² F
frequency of the generator, f = 343 Hz
Since the inductor, and capacitor are connected in series, same current will pass through them.
The current through them is given by;
[tex]I_{rms} = \frac{V_c_{(rms)}}{X_c} = \frac{V_l_{(rms)}}{X_L}[/tex]
where;
Xc and XL are capacitive and inductive reactance respectively;
[tex]\frac{V_c_{(rms)}}{\frac{1}{2\pi fC} } = \frac{V_l_{(rms)}}{2\pi fL}\\\\\frac{V_c_{(rms)}(2\pi fC)}{1} = \frac{V_l_{(rms)}}{2\pi fL}\\\\V_l_{(rms)} = V_c_{(rms)}(2\pi fC)(2\pi fL)\\\\V_l_{(rms)} =V_c_{(rms)}(4\pi^2 f^2 CL)\\\\V_l_{(rms)} = (2.42)(4\pi^2)(343)^2(5.27*10^{-12})(85.2*10^{-3})\\\\V_l_{(rms)} = 5.05 *10^{-6} \ V[/tex]
Therefore, the rms voltage across the inductor is 5.05 x 10⁻⁶ V.
