For the given circuit,the currents I and I_2 | vedclass

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the loop $ABD$: $2I_1 + I_2(G) - 3(I - I_1) = 0$. Assuming $G = 1 \, \Omega$ as per standard bridge pr

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$\frac{70}{24}, \frac{5}{12} \, A$

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(C) Applying Kirchhoff's Voltage Law $(KVL)$ to the loop $ABD$: $2I_1 + I_2(G) - 3(I - I_1) = 0$. Assuming $G = 1 \, \Omega$ as per standard bridge problems of this type,the equation becomes $2I_1 + I_2 - 3I + 3I_1 = 0 \Rightarrow 5I_1 + I_2 = 3I \dots(1)$ Applying $KVL$ to loop $BCD$: $3(I_1 - I_2) - 2(I - I_1 + I_2) - I_2 = 0 \Rightarrow 3I_1 - 3I_2 - 2I + 2I_1 - 2I_2 - I_2 = 0 \Rightarrow 5I_1 - 6I_2 = 2I \dots(2)$ Subtracting $(2)$ from $(1)$: $(5I_1 + I_2) - (5I_1 - 6I_2) = 3I - 2I \Rightarrow 7I_2 = I \Rightarrow I_2 = I/7$. Substituting $I = 7I_2$ into $(1)$: $5I_1 + I_2 = 3(7I_2) \Rightarrow 5I_1 = 20I_2 \Rightarrow I_1 = 4I_2$. Applying $KVL$ to the outer loop: $10 - 1(I) - 2I_1 - 3(I_1 - I_2) = 0$ $10 - 7I_2 - 2(4I_2) - 3(4I_2 - I_2) = 0$ $10 - 7I_2 - 8I_2 - 9I_2 = 0 \Rightarrow 10 = 24I_2 \Rightarrow I_2 = 10/24 = 5/12 \, A$. Then $I = 7 	imes (5/12) = 35/12 \, A$. Note: The provided options suggest $I = 70/24 = 35/12 \, A$ and $I_2 = 5/12 \, A$.

For the given circuit,the currents $I$ and $I_2$ are respectively:

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