Question

Two coils P and S have a mutual inductance of (\(π\)) mH. The secondary coil S has resistance \(4\,\Omega\) and self inductance \((60/π)\) mH. If the current in the primary is \(I_p = 12sin(50πt)\), then the maximum value of the current induced in coil S is [Take \(π^2 = 10\)]

• A. \(2\) A
• B. \(1.8\) A
• C. \(1.5\) A
• D. \(1.2\) A

Hint:

The induced current is limited by the impedance of the secondary, which includes its inductive reactance.

Answer 1

The Correct Option is D

Step 1: Two steps
First find the peak emf from mutual induction, then divide by the impedance.

Step 2: Peak emf
$M\,\omega I_0 = (\pi\times10^{-3})(50\pi)(12) = 0.6\pi^2 = 6$ V using $\pi^2 = 10$.

Step 3: Impedance
$\omega L = 3\ \Omega$ and $R = 4\ \Omega$ form a 3-4-5 triangle, so $Z = 5\ \Omega$.

Step 4: Result
$I_{max} = \dfrac{6}{5} = 1.2$ A.

Final Answer:
The peak current in S is 1.2 A. This is option (D). \[ \boxed{\text{(D) }1.2\ \text{A}} \]