a) Write a set of mesh-current equations that describe the circuit in Fig. 6.25 in terms of the currents i_{1} and i_{2}
b) Verify that if there is no energy stored in the circuit at t=0 and if i_{g}=16-16e^{-5t} A, the solutions for i_{1} and i_{2} are
i_{1}=4+64e^{-5t}-68e^{-4t} A,
i_{2}=1-52e^{-5t}+51e^{-4t} A.
a) Summing the voltages around the i_{1} mesh yields
4\frac{di_{1}}{dt}+8\frac{d}{dt}\left(i_{g}-i_{2}\right) +20 \left(i_{1}-i_{2}\right)+5\left(i_{1}-i_{g}\right)=0.
The i_{2} mesh equation is
20 \left(i_{2}-i_{1}\right)+60 i_{2}+16\frac{d}{dt}\left(i_{2}-i_{g}\right)-8\frac{di_{1}}{dt}=0.
Note that the voltage across the 4 H coil due to the current \left(i_{g}-i_{2}\right) that is, \frac{8d\left(i_{g}-i_{2}\right)}{dt},is a voltage drop in the direction ofi_{1} The voltage induced in the 16 H coil by the current i_{1}that is,8\frac{di_{1}}{dt} ,is a voltage rise in the direction of i_{2}.
b) To check the validity of i_{1} and i_{2} , we begin by testing the initial and final values of i_{1} and i_{2}. We know by hypothesis that i_{1}\left(0\right) = i_{2} \left(0\right)=0. From the given solutions we have
i_{1}\left(0\right)=4 + 64 – 68 = 0,
i_{2}\left(0\right)= 1 – 52 + 51 = 0.
Now we observe that as t approaches infinity the source current\left(i_{g}\right) approaches a constant value of 16 A, and therefore the magnetically coupled coils behave as short circuits. Hence at t = ∞ the circuit reduces to that shown in Fig. 6.26. From Fig. 6.26 we see that at t = ∞ the three resistors are in parallel across the 16 A source. The equivalent resistance is 3.75Ω and thus the voltage across the 16 A current source is 60 V. It follows that
i_{1}\left(\infty \right)=\frac{60}{20} + \frac{60}{60}=4 A,
i_{2}\left(\infty \right)= \frac{60}{60}=1 A.
These values agree with the final values predicted by the solutions for i_{1} and i_{2} Finally we check the solutions by seeing if they satisfy the differential equations derived in (a). We will leave this final check to the reader via Problem 6.37.
