In the circuit shown below E_1 = 4.0 , V,R_1 | vedclass

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(B) Applying Kirchhoff's voltage law for the loops $(1)$ and $(2)$ as shown in the figure:For loop $(1)$:$-2 i_1 - 2(i_1 - i_2) + 4 = 0$$\Rightarrow 4

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$1.8$

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(B) Applying Kirchhoff's voltage law for the loops $(1)$ and $(2)$ as shown in the figure:For loop $(1)$:$-2 i_1 - 2(i_1 - i_2) + 4 = 0$$\Rightarrow 4 i_1 - 2 i_2 = 4$$\Rightarrow 2 i_1 - i_2 = 2$ ......... $(i)$For loop $(2)$:$-2(i_2 - i_1) - 4 i_2 + 6 = 0$$\Rightarrow 2 i_1 - 6 i_2 = -6$$\Rightarrow i_1 - 3 i_2 = -3$ ......... $(ii)$Solving equations $(i)$ and $(ii)$:From $(i)$,$i_2 = 2 i_1 - 2$.Substituting into $(ii)$:$i_1 - 3(2 i_1 - 2) = -3$$i_1 - 6 i_1 + 6 = -3$$-5 i_1 = -9$$i_1 = 1.8 \, A$.

In the circuit shown below $E_1 = 4.0 \, V$,$R_1 = 2 \, \Omega$,$E_2 = 6.0 \, V$,$R_2 = 4 \, \Omega$ and $R_3 = 2 \, \Omega$. The current $I_1$ is ............... $A$

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