Question

4. (25 pt) Two electrical circuits interact magnetically, by "mutual induction." Each circuit is responsive to the changing magnetic fields generated by time- dependent currents in the other. The mutual inductance M quantifies the strength of the interaction. Ci し2 C2 Figure 0.1: Two LC circuits, coupled by a mutual inductance M (a). (5 pt) Through the mutual inductance M, an extra voltage is induced on the inductor, - MdI/dt, where the current is carried by the other circuit. Show that for the circuit above, you have the following equations: (b). (5 pt) Solve for the normal (or eigen) frequencies of the circuit. (c). (10 pt) If the capacitor C1 in the circuit on the right starts with q 1 C and qi = 0, find qi (t) and q2(t), for C,-C2 = 1μΕ, Li = L2 = 10 m H, and M = 1 mH Plot the charges as a function of time. Do you see the beating phenomena, i.e., the energy transftering between the two circuits? If so, what is the beating frequency? (d). (5 pt) If there is a finite resistance R-1 Ω in series in each crtuit. Find q1 (t) and q2(t). Plot the charges as a function of time. What is the charge decay rate, as a result of the resistive dissipation?

Answer 1

"Two electrical circuits interact magnetically by ""mutual induction"" each circuit is responsible to the changing magnetic fields generated by the time dependent current in the other Let I_1 and I_2 be currents in two circuit as shown in above fig. (a) through the mutual induction M, an extra voltage is induced on the inductor. Ans: - Given E = M dI/dt where the current is carried by the other circuit. To show: - L_1 q_1 + Mq_2 + q_1/c_1 = 0 L_2 q_2 + M q_1 + q_2/c_2 = 0 The potential induced due to second circuit (inductor) in the first due to mutual inductance is = -M dI_2/dt = -M q_2 In first circuit Kirchoff�s law along the loop - M di_2/dt - L dI_1/dt - q_1/c_1 = 0 q_1/c_1 + L_1 d^2 q_1/dt^2 + M d^2 q_2/sdt^2 = 0 q_2/c_2 + L_1 q_2 + M q_2 = 0"