Challenging Electricity Problem!

Resistors with no current to flow are having potential difference of 0 V and therefore can be detached from the circuit. This simplify the whole circuit into resultant resistances of:

(3 $\Omega$ ~ 1 $\Omega$ ~ r $\Omega$ ) // (1 $\Omega$ ~ $\frac{4}{3}$ $\Omega$ )

Further simplified into:

(4 $\Omega$ ~ r $\Omega$ ) // ( $\frac73$ $\Omega$ )

Voltage across 2 $\Omega$ resistor with 3 A is 6 V. Another resultant resistance of 2 $\Omega$ in series with it formed from 3 $\Omega$ // 6 $\Omega$ is also 6 V. Therefore, voltage across 4 $\Omega$ resistor is 12 V.Since 4 $\Omega$ , 2 $\Omega$ , 6 $\Omega$ and 3 $\Omega$ resulted with 2 $\Omega$ , in series with 3 $\Omega$ , 3 $\Omega$ and 3 $\Omega$ in parallel of 1 $\Omega$ as resultant resistance, voltage across (1 $\Omega$ ~ 1 $\Omega$ ) // 2 $\Omega$ is therefore 18 V.

For semi-resultant of (3 $\Omega$ ~ 1 $\Omega$ ~ r $\Omega$ ) // (1 $\Omega$ ~ $\frac{4}{3}$ $\Omega$ ), total current through the 2nd 1 $\Omega$ resistor is therefore 18 A.

To (4 $\Omega$ ~ r $\Omega$ ) // ( $\frac73$ $\Omega$ ), the later is having voltage of $\frac73 \Omega \times 18 A = 42 V$

Note that r = 3 $\Omega$ as 42 V - 4 $\Omega \times$ 6 A = 18 V, and therefore $\frac{18 V}{6 A} = 3 \Omega.$ The whole circuit draws an amount of 24 A which is a high loading to usual batteries. The batteries must be high power batteries to be able to give a voltage required.

Voltage across the batteries or cell, E = 42 V

Answer: $\boxed{42}$