In the given circuit, the values of I_1, I_2, | vedclass

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(A) Let the nodes be $A, B, C, D$ as labeled in the figure.Applying Kirchhoff's Voltage Law $(KVL)$ in loop $ABDA$:$2I_1 + 4 - 1I_2 = 0 \implies I_2 =

Vedclass · Accepted Answer

$1.364 	ext{ A}, 6.727 	ext{ A}, 5.909 	ext{ A}$

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(A) Let the nodes be $A, B, C, D$ as labeled in the figure.Applying Kirchhoff's Voltage Law $(KVL)$ in loop $ABDA$:$2I_1 + 4 - 1I_2 = 0 \implies I_2 = 2I_1 + 4$ ...$(i)$Applying $KVL$ in loop $BCDB$:$1(I_1 + I_3) - 4(I_2 - I_3) - 4 = 0$$I_1 + 5I_3 - 4I_2 = 4$Substituting $I_2$ from $(i)$:$I_1 + 5I_3 - 4(2I_1 + 4) = 4 \implies -7I_1 + 5I_3 = 20 \implies I_3 = \frac{20 + 7I_1}{5}$ ...$(ii)$Applying $KVL$ in the outer loop $ADCA$:$1I_2 + 4(I_2 - I_3) = 10 \implies 5I_2 - 4I_3 = 10$ ...$(iii)$Substituting $(i)$ and $(ii)$ into $(iii)$:$5(2I_1 + 4) - 4\left(\frac{20 + 7I_1}{5}ight) = 10$$50I_1 + 100 - 80 - 28I_1 = 50 \implies 22I_1 = 30 \implies I_1 = \frac{15}{11} \approx 1.364 	ext{ A}$Using $I_1$ in $(i)$: $I_2 = 2\left(\frac{15}{11}ight) + 4 = \frac{74}{11} \approx 6.727 	ext{ A}$Using $I_1$ in $(ii)$: $I_3 = \frac{20 + 7(15/11)}{5} = \frac{65}{11} \approx 5.909 	ext{ A}$

In the given circuit, the values of $I_1, I_2, I_3$ are respectively:

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