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

Explanation

Potential difference =6V
Resistance of resistors `=1Omega` and `2Omega`
Power used through resistors of `2Omega`=?
since, resistors are connected in series thus, total effective resistance `R=1Omega+2Omega+2Omega=3Omega`
We know : `I=V//R=6V-:3Omega=2A`
Since current remains same when resistors are connected in series, hence current through resistor of `2Omega=2A`
thus, power (P) `=I^2xxR(2A)^2xx2Ohm=8W`
A 4V battery in parallel with `12Omega` and `2Omega` resistors
Potential difference, V=4V
Resistance of resistors connected in parallel `=12Omega` and `2Omega`
Power used by resistors having resistance of `2Omega`=?
Since, voltage across the circuit remains same if resistor are connected in parallel
Thus, power(P) used by resistance of 2 Ohm
`=V^2//R=(4V)^2//2Omega=16//2=8W`
Thus power used by resistance of `2Omega` in both the case =8W