AN LIKC CIRCUIT HAS A VARIAALE FCQUENCY POWER SUPNLY WITH V Vs. DRIVING FREQUENCY GRAPH IS. SHOWN. = 20V. ITS CURRENT RMS IRMSLA) 15 THE RMS CURRGNT AMPLITUDG. IF THe INDUCTANLE IN THE CIRCUIT 13 3 HENKIGS, FIND: fCH2) 1. 500 a) b)c c) Xe if 2) X IF f = S0OH2 e) Z f= SOOHZ (1

Solution:

Given Values,

$$\begin{gathered} \text{RMS Voltage (}{\mathit{\text{V}}}_{\mathit{R}\mathit{M}\mathit{S}})=20 \mathrm{V} \\ \mathrm{Inductance} (L)=3 \mathrm{H} \end{gathered}$$

From the above Given Current v/s driving frequency graph:

We have,

$$\begin{gathered} \text{RMS current} \left(I_{RMS}\right)=5 \text{A} \\ \text{Resonant frequency} \left(f_r\right)=500 \text{Hz} \end{gathered}$$

So,

The Values of each given parameters are calculated as follows:

a) Resistance (R)

We know that,

At Resonance Condition:

$$\begin{gathered} X_L=X_C \text{which occurs at 500 Hz} \\ \text{and } \\ Z=R (\text{Frequency at resonance for} Z \text{wil be} 0) \end{gathered}$$

So,

Here,

$$\begin{gathered} Z=R=\frac{V_{RMS}}{I_{RMS}} \\ R=\frac{20 V}{5 A}=4 \Omega \\ \text{Thus,} \\ \mathbf{\text{Resistance }}\mathbf{\left(}\mathit{R}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{ }\mathbf{\text{Ω}} \end{gathered}$$

b) Capacitance (C):

We know that,

Capacitive Reactance$\left(X_C\right)=\frac{1}{\omega C}$

Here,

$$\begin{gathered} C=\frac{1}{\omega X_C} \\ C=\frac{1}{\left(2{\mathrm{\pi f}}_{\mathrm{r}}\right)X_C} \end{gathered}$$

Since we know that at resonance condition:

$$\begin{gathered} X_C=X_L \\ {X}_L=\omega L \\ =2\pi f_{r}L \\ =2\times 3.14\times 500\text{ Hz×3 H} \\ =9420 \Omega \\ \text{Hence,} \\ {X}_C=X_L =9420 \Omega \end{gathered}$$

As per the above relation, we get:

$$\begin{gathered} C=\frac{1}{\left(2{\mathrm{\pi f}}_{\mathrm{r}}\right)X_C} \\ =\frac{1}{\left(2\times 3.14\times 500 \text{Hz}\right)\times 9420 \Omega } \\ C=\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathbf{\text{F =33.8 nF}} \\ \text{Thus,} \\ \text{The capacitence Value is}\mathbf{ }\mathbf{ }\mathbf{\text{33.8 nF}} \end{gathered}$$