Question

Consider two coupled circuits, having self-inductances \( L_1 \) and \( L_2 \), that carry non-zero currents \( I_1 \) and \( I_2 \), respectively. The mutual inductance between the circuits is \( M \) with unity coupling coefficient. The stored magnetic energy of the coupled circuits is minimum at which of the following value(s) of \( \frac{I_1}{I_2} \)?

• A. \( -\frac{M}{L_1} \) {2cm}
• B. \( -\frac{M}{L_2} \) {2cm}
• C. \( -\frac{L_1}{M} \) {2cm}
• D. \( -\frac{L_2}{M} \)

Hint:

The magnetic energy in two coupled inductors is: \[ W = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 + M I_1 I_2 \] To minimize \( W \), differentiate with respect to \( \frac{I_1}{I_2} \), set derivative to zero, and solve: \[ \frac{I_1}{I_2} = -\frac{M}{L_1}, \quad {or equivalently} \quad \frac{I_2}{I_1} = -\frac{M}{L_2} \Rightarrow \frac{I_1}{I_2} = -\frac{L_2}{M} \] So, both (A) and (D) are valid.

Answer 1

The Correct Option is A, D

The total magnetic energy stored in two magnetically coupled inductors is given by: \[ W = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 + M I_1 I_2 \] To find the **minimum** energy, consider \( I_1 \) and \( I_2 \) to be variables with a constant ratio: \[ \frac{I_1}{I_2} = k \Rightarrow I_1 = k I_2 \] Substitute into the energy expression: \[ W = \frac{1}{2}L_1 (k I_2)^2 + \frac{1}{2}L_2 I_2^2 + M (k I_2)(I_2) = \frac{1}{2}L_1 k^2 I_2^2 + \frac{1}{2}L_2 I_2^2 + M k I_2^2 \] Factor out \( I_2^2 \): \[ W = I_2^2 \left( \frac{1}{2}L_1 k^2 + \frac{1}{2}L_2 + M k \right) \] To minimize \( W \), minimize the expression inside the brackets: \[ f(k) = \frac{1}{2}L_1 k^2 + M k + \frac{1}{2}L_2 \] Differentiate and set to zero: \[ \frac{df}{dk} = L_1 k + M = 0 \Rightarrow k = -\frac{M}{L_1} \Rightarrow \frac{I_1}{I_2} = -\frac{M}{L_1} \] Alternatively, we can express it in terms of \( \frac{I_2}{I_1} = -\frac{M}{L_2} \Rightarrow \frac{I_1}{I_2} = -\frac{L_2}{M} \) So both options (A) and (D) are correct. \[ \boxed{{Correct options: (A), (D)}} \]