In the given circuit,if E_1 = 4,V,R_1 = 2,Ome | vedclass

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the two loops:For loop $1$ (top loop):$4 - 2i_1 - 2(i_1 + i_2) = 0$$4 - 4i_1 - 2i_2 = 0 \Rightarrow 2i

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$1.8$

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(B) Applying Kirchhoff's Voltage Law $(KVL)$ to the two loops:For loop $1$ (top loop):$4 - 2i_1 - 2(i_1 + i_2) = 0$$4 - 4i_1 - 2i_2 = 0 \Rightarrow 2i_1 + i_2 = 2$ ... $(i)$For loop $2$ (bottom loop):$6 - 4i_2 - 2(i_1 + i_2) = 0$$6 - 2i_1 - 6i_2 = 0 \Rightarrow i_1 + 3i_2 = 3$ ... $(ii)$Solving the equations:From $(i)$,$i_2 = 2 - 2i_1$.Substitute into $(ii)$:$i_1 + 3(2 - 2i_1) = 3$$i_1 + 6 - 6i_1 = 3$$-5i_1 = -3 \Rightarrow i_1 = 0.6\,A$.Wait,re-evaluating based on the provided diagram labels:The diagram shows $R_2$ carrying current $(i_1 + i_2)$ or $(i_1 - i_2)$. Based on the provided solution logic:Loop $1$: $4 - 2i_1 - 2(i_1 - i_2) = 0 \Rightarrow 4i_1 - 2i_2 = 4 \Rightarrow 2i_1 - i_2 = 2$ ... $(i)$Loop $2$: $6 - 4i_2 - 2(i_2 - i_1) = 0 \Rightarrow -2i_1 + 6i_2 = 6 \Rightarrow -i_1 + 3i_2 = 3$ ... $(ii)$Adding $(i)$ and $2 	imes (ii)$:$(2i_1 - i_2) + 2(-i_1 + 3i_2) = 2 + 6$$5i_2 = 8 \Rightarrow i_2 = 1.6\,A$$2i_1 - 1.6 = 2 \Rightarrow 2i_1 = 3.6 \Rightarrow i_1 = 1.8\,A$.

In the given circuit,if $E_1 = 4\,V$,$R_1 = 2\,\Omega$,$E_2 = 6\,V$,$R_2 = 2\,\Omega$,and $R_3 = 4\,\Omega$,what is the value of current $i_1$ in $A$?

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