In given circuit `R_(1)=5Omega,R_(2)=3Omega=7Omega` and supply voltage is 10V. Calculate power dissipated in `R_(1)` and `R_(3)`.

Explanation

As voltage across `R_(1)` is 10V so power dissipated in `R_(1) P_(1)=(V^(2))/(R_(1))=(100)/(5)=20W`
As `R_(2)` and `R_(3)` are connected in series so current in Power dissipated in `R_(3),P_(3)=I ^ (2)R_(3)=1xx7=7W R_(3)=(V)/(R_(2)+R_(3))=(10)/(3+7)=1A`