A 5 H inductor `80 Omega` resistor and `3muF ` capacitor are connected in series with an AC source with variable frequency such that the value of current is maximum in the circuit . Calculate the value of inductive reactance capacitive reactance and total impedance . Also calculate the value of peak current if value of peak emf of source is 220 V.

Answer

To solve the problem step by step, we need to calculate the inductive reactance (XL), capacitive reactance (XC), total impedance (Z), and the peak current (I0) in the circuit.

### Step 1: Calculate the Angular Frequency (ω)
Since the current is maximum when the inductive reactance (XL) equals the capacitive reactance (XC), we can express this condition mathematically.

The formula for angular frequency (ω) is given by:
\[ \omega = \frac{1}{\sqrt{L \cdot C}} \]

Where:
- \(L = 5 \, H\) (inductance)
- \(C = 3 \, \mu F = 3 \times 10^{-6} \, F\)

Substituting the values:
\[ \omega = \frac{1}{\sqrt{5 \cdot 3 \times 10^{-6}}} \]
\[ \omega = \frac{1}{\sqrt{15 \times 10^{-6}}} = \frac{1}{\sqrt{0.000015}} \approx 258.2 \, rad/s \]

### Step 2: Calculate Inductive Reactance (XL)
Inductive reactance is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 258.2 \cdot 5 = 1291 \, \Omega \]

### Step 3: Calculate Capacitive Reactance (XC)
Capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{258.2 \cdot 3 \times 10^{-6}} \approx 1291 \, \Omega \]

### Step 4: Calculate Total Impedance (Z)
Since \(X_L = X_C\), the total impedance in the circuit is simply the resistance:
\[ Z = R = 80 \, \Omega \]

### Step 5: Calculate Peak Current (I0)
The peak current can be calculated using Ohm's law:
\[ I_0 = \frac{V_0}{Z} \]
Where \(V_0 = 220 \, V\):
\[ I_0 = \frac{220}{80} = 2.75 \, A \]

### Summary of Results:
- Inductive Reactance (XL) = 1291 Ω
- Capacitive Reactance (XC) = 1291 Ω
- Total Impedance (Z) = 80 Ω
- Peak Current (I0) = 2.75 A