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 ..................... $A$
Diffcult
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(b) Applying Kirchhoff’s law for the loops $(1) $ and $(2)$ as shown in figure
For loop $(1)$
$ - 2{i_1} - 2({i_1} - {i_2}) + 4 = 0$ $==>$ $2{i_1} - {i_2} = 2$ …$(i)$
For loop $(2)$
$ - 2({i_1} - {i_2}) + 4{i_2} - 6 = 0$ $==>$ $ - {i_1} + 3{i_2} = 3$ …$(ii)$
On solving equation $(i)$ and $(ii) $ ${i_1} = 1.8\,A$.
For loop $(1)$
$ - 2{i_1} - 2({i_1} - {i_2}) + 4 = 0$ $==>$ $2{i_1} - {i_2} = 2$ …$(i)$
For loop $(2)$
$ - 2({i_1} - {i_2}) + 4{i_2} - 6 = 0$ $==>$ $ - {i_1} + 3{i_2} = 3$ …$(ii)$
On solving equation $(i)$ and $(ii) $ ${i_1} = 1.8\,A$.
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