Question

Two inductors (4 H with current \(i_1(t)=1\sin(200\pi t)\,\text{mA}\) and 5 H with \(i_2(t)=2\sin(200\pi t)\,\text{mA}\)) are mutually coupled with coupling coefficient \(k=0.6\). Currents enter the dotted ends. Find the {peak} energy stored in the coupled system (in \(\mu\)J).

Hint:

For two coupled coils, \(W=\frac12L_1i_1^2+\frac12L_2i_2^2\pm Mi_1i_2\); use \(+\) if currents enter the dotted ends (aiding), \(−\) if they leave/enter oppositely.

Answer 1

Correct Answer: 17

Step 1: Mutual inductance. \(M=k\sqrt{L_1L_2}=0.6\sqrt{4\times 5}=2.6833~\text{H}\).
Step 2: Peak currents. Since the sinusoids are in phase, \(I_{1,\text{pk}}=1~\text{mA}=1\times10^{-3}\text{ A}\), \(I_{2,\text{pk}}=2~\text{mA}=2\times10^{-3}\text{ A}\).
Step 3: Peak energy (aiding polarity). \[ W_{\max}=\tfrac12 L_1 I_1^2+\tfrac12 L_2 I_2^2+M I_1 I_2 =\tfrac12(4)(10^{-3})^2+\tfrac12(5)(2\!\times\!10^{-3})^2 +2.6833(10^{-3})(2\!\times\!10^{-3}) \] \[ =2.0\times10^{-6}+1.0\times10^{-5}+5.3666\times10^{-6} =1.7367\times10^{-5}\text{ J}=17.37~\mu\text{J}. \] Final Answer: 17.37 \(\mu\)J