Two resistors of 2 Ω and 6 Ω are connected in series and this combination is connected across a 12 V battery. Find the current in the 6 Ω resistor.

Correct option is A

When resistors are connected in series, the same current flows through each component because there is only one continuous path for current to follow. In this problem, the total resistance $$R_\text{total}$$​ is the sum of the individual resistances:
$$R_\text{total} = R_1 + R_2 = 2\,\Omega + 6\,\Omega = 8\,\Omega.$$​
The voltage source is 12 V, so by Ohm’s Law
(V=IR), the current III through the series combination is calculated as:
$$I = \frac{V}{R_\text{total}} = \frac{12}{8} = 1.5\,\text{A}.$$​