Interpret the circuit

The two cells are connected in parallel between points $A$ and $B$, and the resistor $R=4\,\Omega$ is also connected across $A$ and $B$.

So, between the same two nodes $A$ and $B$, we have three parallel branches:

• Cell 1: emf $E_1=12\,\text{V}$, internal resistance $r_1=3\,\Omega$
• Cell 2: emf $E_2=6\,\text{V}$, internal resistance $r_2=6\,\Omega$
• External resistor: $R=4\,\Omega$

Let the potential difference across $A$ and $B$ be $V$

Then current through the resistor is $I_R=\frac{V}{R}=\frac{V}{4}.$

For each cell branch, taking current from $A$ to $B$ as positive:

• For cell 1, $I_1=\frac{E_1-V}{r_1}=\frac{12-V}{3}.$
• For cell 2, $I_2=\frac{E_2-V}{r_2}=\frac{6-V}{6}.$

Apply Kirchhoff's Current Law at node $A$

The current supplied by the two cells goes through the resistor: $I_1+I_2=I_R.$

Substitute the expressions: $\frac{12-V}{3}+\frac{6-V}{6}=\frac{V}{4}.$

Solve for $V$

Take LCM $=12$: $4(12-V)+2(6-V)=3V.$

Simplify left side: $48-4V+12-2V=3V$ $60-6V=3V$ $60=9V$ $V=\frac{60}{9}=\frac{20}{3}\,\text{V}.$

Find current through $R$

$I_R=\frac{V}{4}=\frac{20/3}{4}=\frac{20}{12}=\frac{5}{3}\,\text{A}.$

Final answer

The current through the resistor is $\boxed{\frac{5}{3}\text{ A}}.$

Since this is an integer-type question, the exact value is $\boxed{\frac{5}{3}}$, not $1$.

Comparison with stored answer

Stored correct answer: $1$

Derived answer: $\dfrac{5}{3}$

These do not match. The stored answer appears incorrect if the circuit is interpreted as the standard parallel connection of the two cells and the resistor across the same terminals $A$ and $B$.