Two resistors of `4Omega` and `6Omega` are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate (i) the power supplied by the battery, (ii) the power dissipated in each resistor.

Explanation

Here `R_(1)= 4Omega` and `R_(2)= 6Omega`
Both are in parallel. Hence
`(1)/(R)= (1)/(R_(1))+ (1)/(R_(2))= (R_(2)+R_(1))/(R_(1)R_(2))`
`R= (R_(1)R_(2))/(R_(2)+R_(1))`
`= (4 xx 6)/(4+6)`
`= 2.4 Omega`
(i) Power dissipated, `P= (V^(2))/(R)= (36 xx 10)/(24)= 15W`
(ii) For `R_(1)`, Power dissipated
`P_(1)= (V^(2))/(R)= (36)/(4)= 9W`
`P_(2)= (V^(2))/(R)= (6 xx 6)/(6)= 6W`