Question

-There is no energy stored in the circuit shown at the time the switch is opened. if k (coupling coefficient) = 0.5, L₁= 4H, L₂ = 9H, R_o =12$\Omega$

a) What is the mutual inductance of the circuit?

b)If i_g = -2e^-t - 2e^-2t - 2 and i₂ = 2e^-t - 2e^-2t , Find the expression for the voltage across the current source v₁ using KVL.

Answer 1

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Given circuit is Mutual Inductance = K squareroot L_1 L_2 = 0.5 times squareroot 4 times 9 = 0.5 times 6 = 3H Given i_g = -2e^-t - 2e^-2t - 2 & i_2 = 2e^-t - 2e^-2t By taking i_g & i_2 into consideration with their directions and values V_1 can be written as V_1 = -(-L_1 di_g/dt + M di_2/dt) = -(4 times d/dt (+ 2e^-t + 2e^-2t -2) + 3 times d/dt (2e^-t - 2e^-2t)) = -(4 times (-2e^-t -4e^-2t) + 3 (-2e^-t + 4 e^-2t) V_1 = 2e^-t + 28 e^-2t V