Compare the power used in the `2Omega` resistor in each of the following circuits:
i] a 6 V battery in series with `1Omegaand2Omega` resistors, and
ii] a 4 V battery in parallel with `12Omegaand2Omega` resistors.

Explanation

Case [i]
Current in the circuit,
`I=V/(R_(1)+R_(2))=6/3=2A`
Power used = `I^(2)R=(2)^(2)xx2=2xx2xx2=8W`
`R=1+2=3Omega`
`I=V/R=6/3=2A`
Therefore Voltage across `2Omega` resistor = `1xx2=2xx2=4V`
`PI=VI=4xx2=8W`
Case [ii] `I_(2)=V/R_(2)=4/2=2A`
`P_(2)` across `2Omega` resistor = `4xx2=8W`
Therefore ratio, `P_(1):P_(2)=8:8=1`