In the circuit shown : $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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$\mathrm{V}_{\mathrm{AB}}=\frac{\frac{4}{2}+0-\frac{6}{4}}{\frac{1}{2}+\frac{1}{2}+\frac{1}{4}}-\frac{2}{5}=0.4 \mathrm{\,V}$
$\mathrm{V}_{\mathrm{R}_{1}}=\mathrm{E}_{1}-\mathrm{V}_{\mathrm{AB}}=3.6 \mathrm{\,V}$
$\mathrm{i}_{\mathrm{R}_{1}}=\frac{\mathrm{V}_{\mathrm{R}_{1}}}{\mathrm{R}_{1}}=\frac{3.6}{2}=1.8 \mathrm{\,A}$
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