A 50 mH inductor is connected across a power supply of emf 130 sin (314 t) volts. Calculate the resistance of the inductor if maximum current in the circuit is 8 A.

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify Given Values
- Inductance (L) = 50 mH = 50 × 10^(-3) H
- EMF (V) = 130 sin(314t) volts
- Maximum Current (I₀) = 8 A
- Angular frequency (ω) = 314 rad/s (from the equation of EMF)

### Step 2: Calculate Inductive Reactance (Xₗ)
Inductive reactance (Xₗ) is given by the formula:
\[ X_L = L \cdot \omega \]
Substituting the values:
\[ X_L = 50 \times 10^{-3} \, H \times 314 \, \text{rad/s} \]
Calculating this:
\[ X_L = 15.7 \, \Omega \]

### Step 3: Calculate Maximum Voltage (V₀)
The maximum voltage (V₀) can be taken directly from the EMF equation:
\[ V_0 = 130 \, V \]

### Step 4: Calculate Impedance (Z)
The impedance (Z) of the circuit can be calculated using the formula:
\[ Z = \frac{V_0}{I_0} \]
Substituting the values:
\[ Z = \frac{130 \, V}{8 \, A} \]
Calculating this:
\[ Z = 16.25 \, \Omega \]

### Step 5: Relate Impedance, Resistance (R), and Inductive Reactance (Xₗ)
The relationship between impedance (Z), resistance (R), and inductive reactance (Xₗ) in an RL circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the known values:
\[ 16.25 = \sqrt{R^2 + (15.7)^2} \]

### Step 6: Solve for Resistance (R)
Squaring both sides:
\[ (16.25)^2 = R^2 + (15.7)^2 \]
Calculating the squares:
\[ 264.0625 = R^2 + 246.49 \]
Rearranging to find R²:
\[ R^2 = 264.0625 - 246.49 \]
Calculating:
\[ R^2 = 17.5725 \]
Taking the square root:
\[ R = \sqrt{17.5725} \approx 4.19 \, \Omega \]

### Final Answer
The resistance of the inductor in the circuit is approximately **4.19 Ω**.

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