Compare the power used in the `2 Omega` resistor in each of the following circuits :
(i) a `6V` battery in series with `1 Omega and 2 Omega` resistors, and
(ii) a `4 V` battery in parallel with `12 Omega and 2 Omega` resistors.

Explanation

(i) In the First Case:
Potential difference, `V = 6V`
Current, `I = ?` (To be calculated)
And, Resistance, `R = 1 Omega + 2 Omega` (Resistors in series)
`R = 3 Omega`
Now, by Ohm's law (for whole circuit) : `(V)/(I) = R`
So, `(6)/(I) = 3`
And, Current, `I = (6)/(3)`
`I = 2A`
Thus, the current flowing in the circuit containing `1 Omega` and `2 Omega` resistors in series is 2 A. In series circuit, the same current flows throughout the circuit. So, the current flowing through the `2 Omega` resistor is also 2A [Figure (i)].Now, we know that the current (in `2 Omega` resistor) is 2A and its resistance is `2 Omega`.

So, Power used in `2 Omega` resistor, `P_(1) = I^(2) xx R`
(In first case) `P_(1) = (2)^(2) xx 2`
`P_(1) = 4 xx 2`
`P_(1) = 8W` ...(1)
Thus, the power `P_(1)` used in `2 Omega` resistor in the first case is 8 watts.
(ii) In the Second Case:
Here 4 V battery is attached across the parallel combination of `12 Omega` and `2 Omega` resistors, so the potential difference across the `2 Omega` resistor will also be 4V [see Figure (ii)].
Now, Potentil difference, `V = 4V` (Across the `2 Omega` resistor)
And, Resistance, `R = 2 Omega`
So, Power used in `2 Omega` resistor, `P_(2) =(V^(2))/(R)`
`P_(2) = ((4)^(2))/(2)`
`P_(2) = (16)/(2)`
`P_(2) = 8W` ...(2)
Thus, the power `P_(2)` used in the `2 Omega` resistor in the second case is also 8 watts.
In order to compare the power used in `2 Omega` resistor in the two circuits, let us divide `P_(1)` by `P_(2)`.
So, `(P_(1))/(P_(2)) = (8W)/(8W)`
or `(P_(1))/(P_(2)) = 1`
or `P_(1) = P_(2)`
The `2 Omega` resistor uses equal power in both the circuits.