For the circuit shown, with $R_{1} = 1.0\Omega, R_{2} = 2.0\Omega, E_{1} = 2\ V$ and $E_{2} = E_{3} = 4\ V$, the potential difference between the points $'a'$ and $'b'$ is approximately (in $V$).

$E_{eq} = \dfrac {\dfrac {E_{1}}{2R_{1}} + \dfrac {E_{2}}{R_{2}} + \dfrac {E_{3}}{2R_{1}}}{\dfrac {1}{2R_{1}} + \dfrac {1}{R_{2}} + \dfrac {1}{2R_{1}}}$

$= \dfrac {\dfrac {2}{2} + \dfrac {4}{2} + \dfrac {4}{2}}{\dfrac {1}{2} + \dfrac {1}{2} + \dfrac {1}{2}}$

$= \dfrac {5}{\dfrac {3}{2}} = \dfrac {10}{3} = 3.3$.