The 10 ohm and 15 ohm resistors are connected in parallel to each other but collectively connected in series with the 12 ohm resistor. The EMF of the battery is 18 V.

What we need to do is to first calculate the overall resistance of the circuit to find out the current which flows in the main circuit.

For the 10 ohm and 15 ohm resistors connected in parallel, their effective resistance is given by

1/R = 1/10 + 1/15
1/R = 1/6
R = 6 ohm

Together with the 12 ohm resistance, the overall resistance of the entire circuit is given by

R = 12 + 6
R = 18 ohm

Since the overall resistance is 18 ohm and the EMF of the battery is 18 V, assuming no internal resistance or other complications, current flowing through the full circuit
= voltage / resistance
= 18 / 18
= 1 A

Now, it is noteworthy to know that in series, currents are the same in all segments of the circuit (and equal to the current in the main circuit), but voltage values add up across resistors. Likewise, in parallel, voltage values are the same across branches (and equal to the preceding path splitting the paths into branches) but current values in these branches add up to the preceding path.

What’s going to happen here is that the 12 ohm resistor is in series to the main circuit, so it’s going to take up some potential difference (ie voltage) away from the 18 V EMF of the battery.

Potential difference across the 12 ohm resistor
= current (main circuit) x resistance
= 1 x 12
= 12 V

So, essentially 12 V out of the 18 V available has been consumed by the 12 ohm battery, leaving 6 V behind.

Since the 10 ohm and 15 ohm resistances are in parallel, they will have the same voltage as each other and as the preceding path of 6 V, so the voltage across the 10 ohm resistor will be 6 V and so does the 15 ohm resistor with 6 V (note: the 6 V does not split into 3 V and 3 V)

Since the voltage across the 10 ohm resistor is 6 V, current across the 10 ohm resistor
= voltage / resistance
= 6 / 10
= 0.6 A

——————————

You can check that the values are correct by also considering the following continuation.

Since the voltage across the 15 ohm resistor is also 6 V, current across the 15 ohm resistor
= voltage / resistance
= 6 / 15
= 0.4 A

Because the 10 ohm and 15 ohm resistors are in parallel, the current values add up to the preceding branch (which happens to also be the main branch).

Sum of currents in both branches
= 0.6 + 0.4
= 1.0 A

which is the same in value as the main branch.