Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3 $$\Omega$$. Cell 2 has emf of 6 V and internal resistance of 6 $$\Omega$$. An external resistor R of 4 $$\Omega$$ is connected across A and B. The current flowing through R will be _____ A.

step 1: combine the two cells in parallel (proper way)

For parallel cells:

$$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}$$

But here, from polarity:

• one contributes +12/3
• the other contributes −6/6 (opposing direction)

So:

$$E_{eq}=\frac{12/3-6/6}{1/3+1/6}=\frac{4-1}{1/2}=\frac{3}{1/2}=6\text{ V}$$

step 2: equivalent internal resistance

$$r_{eq}=3∥6=2Ω$$

step 3: final circuit

Now it becomes:

6 V source in series with 2 Ω and external 4 Ω

Total resistance:

Rtotal=2+4=6 Ω

step 4: current

$$I=\frac{6}{6}=1A$$