Question

Two coils A and B have 180 and 360 turns respectively. The current of 1A flows through both the coils. Due to current of 1A in coil A, flux per turn of \(0.8\times 10^{-3}\) Wb is linked with coil A. Due to current of 1A in coil B, flux per turn of \(1\times 10^{-3}\) Wb is linked with coil B. The self inductance of coil A is \(L_A\) and the self inductance of coil B is \(L_B\). The ratio \(L_A\) to \(L_B\) is

• A. \(\frac{1}{5}\)
• B. \(\frac{2}{5}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{5}{2}\)

Hint:

Self inductance L = N times flux per turn divided by current.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The self inductance of a coil is its total flux linkage per unit current: \(L = \dfrac{N\phi}{I}\), where \(\phi\) is the flux per turn.

Step 2: Key Formula or Approach:
Both coils carry 1 A, so \(L = N\phi\) numerically.

Step 3: Detailed Explanation:
Coil A:
\[ L_A = \frac{180\times0.8\times10^{-3}}{1} = 0.144 \text{ H} \]
Coil B:
\[ L_B = \frac{360\times1\times10^{-3}}{1} = 0.36 \text{ H} \]
Ratio:
\[ \frac{L_A}{L_B} = \frac{0.144}{0.36} = 0.4 = \frac25 \]
Option (A) \(\tfrac15\) results from using only the turns ratio \(\tfrac{180}{360}\) times \(\tfrac12\). Option (D) \(\tfrac52\) is the inverse of the correct ratio.

Final Answer:
\(L_A : L_B = 2:5\), option (B). \[ \boxed{\frac{2}{5} \text{ (B)}} \]