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: Understand the concept
The changing current in P induces an emf in S: \(\varepsilon_S = M\dfrac{dI_p}{dt}\). The secondary has resistance and self inductance, so its current is limited by its impedance.

Step 2: Maximum emf
\(I_p = 12\sin(50\pi t)\), so \(\dfrac{dI_p}{dt} = 600\pi\cos(50\pi t)\) with peak \(600\pi\) A/s. With \(M = \pi\times10^{-3}\) H:
\[ \varepsilon_{max} = \pi\times10^{-3}\times600\pi = 0.6\pi^2 = 6\ \text{V} \]

Step 3: Impedance of S
Reactance \(X_L = \omega L = 50\pi\times\dfrac{60}{\pi}\times10^{-3} = 3\ \Omega\). Resistance is \(4\ \Omega\), so \(Z = \sqrt{4^2 + 3^2} = 5\ \Omega\).

Step 4: Maximum current
\[ I_{max} = \frac{6}{5} = 1.2\ \text{A} \]
Option (D). If the self inductance were ignored, we would get \(\frac{6}{4} = 1.5\) A, which is option (C) and is a trap.

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