Consider this circuit:

If we do node voltage analysis at Node A (assuming all currents go out we get)

$$ \frac{V-10}{2R} + \frac{V}{2R} + \frac{V}{2R} + i_1 =0$$

$$ \frac{3V}{2R} – \frac{10}{2R} = -i_1$$

$$ \frac{-1}{2R} (3V-10) = i_1 $$

Now this is where my problem is, I am having trouble expressing i_1 in terms of voltage difference because there is no resistor! I think I might have to use a supernode but I am not quite sure.

## Best Answer

A method that also gives good result is redrawing the sketch in a simpler form as shown below:

Current \$i_3\$ is the sum of \$i_2\$ and \$i_1\$ and equals:

\$\frac{10-5}{2R}=\frac{5}{R}+i_1\$

Solving for \$i_1\$ gives \$i_1=-\frac{2.5}{R}\$. As pointed out by jonk, considering the negative sign, the 5-V generator sources \$i_1\$.