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) potential difference, V=6 V
`1 Omega "and " 2 Omega` resistors are connected in series. Therefore, equivalent resistance of the circuit, `R=1+2=3 Omega`
According to Ohm's law,
V=IR
Where,
I is the current through the circuit
I=6/3=2A
This current will flow through each component of the circuit because there is no division of currnt in series circuits. Henece, Current flowing through the `2Omega` resistor is 2 A. Power is given by the expression.
`P=(I)^(2)R=(2)^(2)xx2=8W`
(ii) Potential difference, V=4 V
`12 Omega "and " 12 Omega` resistors are connected in parallel. The voltage across eath component of a parallel circuit remains the same. Hence, the voltage across `2 Omega` resistor will be 4 V.
Power consumed by `2 Omega` resistor is given by
`P=V^(2)//2=8W`
Therefore, the power used by `2 Omega` resistor is 8 W.