Question 5.13: A current of i = 6(3- e^-2t)A flows in the circuit as shown ......
A current ofi=6(3-\mathrm{e}^{-2t})\mathrm{A} flows in the circuit as shown in Fig. 5.28. Given, i_{2}(0)=4\,\mathrm{A}, calculatei_{1}(0),\,i_{1},\,i_{2} and the voltage across the 3 H inductor.
{\mathrm{At}}\,t=0{\mathrm{,}} i(0)=6(3-1)=12{\mathrm{~A}} (5.210)
i_{1}(0)=i(0)-i_{2}(0)=12-4=8\mathrm{~A} (5.211)
The equivalent inductance is,
L_{\mathrm{eq}}={\frac{(5+3)\times6}{8+6}}=3.43\ \mathrm{H} (5.212)
The source voltage is determined as,
\nu=L_{\mathrm{eq}}{\frac{\mathrm{d}i}{\mathrm{d}t}}=3.43{\frac{\mathrm{d}}{\mathrm{d}t}}6(3-\mathrm{e}^{-2t})=41.14\mathrm{e}^{-2t}\, \mathrm{V} (5.213)
Again, according to the circuit,
\nu_{\mathrm{6H}}=\nu=6{\frac{\mathrm{d}i_{1}}{\mathrm{d}t}} (5.214)
i_{1} =\frac{41.14}{6}\int\limits_{0}^{t}{e^{-2t}dt+i_{1}\left(0\right)} (5.215)
i_{1}=-3.43\mathrm{e}^{-2t}+8\,\mathrm{A} (5.216)
The current i2 can be determined as,
i_{2}=i-i_{1}=6(3-\mathrm{e}^{-2t})+3.43\mathrm{e}^{-2t}-8=10-2.57\mathrm{e}^{-2t}\ \mathrm{A} (5.217)
The voltage across 3 H inductor is,
\nu_{2}=L_{3\mathrm{H}}{\frac{\mathrm{d}i_{2}}{\mathrm{d}t}}=3{\frac{\mathrm{d}}{\mathrm{d}t}}\left(10-2.57\mathrm{e}^{-2t}\right)=15.42\mathrm{e}^{-2t}{\mathrm{~V}} (5.218)