Question

What will be the impedance of a coil having an inductance of 0.1 henry and resistance of 5 ohms to the 50 Hz A.C supply

• A. 31.4 ohm
• B. 31.8 ohm
• C. 25.0 ohm
• D. 25.1 ohm

Hint:

AC Impedance Calculation: $X_L = 2\pi(50)(0.1) = 31.42\;\Omega$. $Z = \sqrt{5^2 + 31.42^2} = \sqrt{25 + 987} = \sqrt{1012} = \mathbf{31.8\;\Omega}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
AC circuit impedance calculations for series R-L coils: total electrical impedance ($Z$) is the vector hypotenuse of pure resistance ($R$) and inductive reactance ($X_L = 2\pi f L$) in the complex impedance plane ($Z = \sqrt{R^2 + X_L^2}$).
Key Formula or Approach:
\[ X_L = 2 \pi f L = 2 \times 3.1416 \times 50\text{ Hz} \times 0.1\text{ H} = \mathbf{31.416 \; \Omega} \]
\[ \mathbf{Z} = \sqrt{R^2 + X_L^2} = \sqrt{(5)^2 + (31.416)^2} = \sqrt{25 + 986.97} = \sqrt{1011.97} = \mathbf{31.81 \; \Omega \approx 31.8 \; \Omega} \]

Step 2: Detailed Explanation:
Step-by-step mathematical calculation of AC coil impedance:
1. Given Data:
- Resistance: $R = 5\;\Omega$
- Inductance: $L = 0.1\text{ Henry (H)}$
- Supply Frequency: $f = 50\text{ Hz}$
2. Calculate Inductive Reactance ($X_L$):
\[ X_L = 2 \pi f L = 2 \times 3.1416 \times 50 \times 0.1 = 31.416\;\Omega \]
3. Calculate Total Impedance ($Z$):
\[ Z = \sqrt{R^2 + X_L^2} = \sqrt{5^2 + (31.416)^2} = \sqrt{25 + 986.97} = \sqrt{1011.97} = \mathbf{31.81\;\Omega \approx 31.8\;\Omega} \]

Step 3: Final Answer:
Hence, the impedance of the coil is 31.8 ohm, matching option (B).