The 10 Ω resistor in FIGURE EX28.29 is dissipating $40W$of power. How much power are the other two resistors dissipating?

We need to find that How much power are the other two resistors dissipating?

$40W$of the power is dissipated by $10\Omega$resistor. So, calculating the current through the resistance $10\Omega$. The energy is dissipated when the current flows through a resistor. The rate of the dissipated energy is the power. This rate where the energy is transferred from the current to the resistor is

$P_R=I^{2}R$ (1)

Use equation (1) and solve it for the current and plug the value for $R=10\Omega$and $P_R=40W$to get $I_{10\Omega }$by

$I_{10\Omega }=\sqrt{\frac{P_R}{R}}=\sqrt{\frac{40W}{10\Omega }}=2A$

The current is the same for the resistors, and as the two resistors $10\Omega$and $5\Omega$are in series, the current $I_{\mathit{10}\mathit{\Omega }}\mathit{}$and $I_{5\Omega }$is same.

$I_{10\Omega }=I_{5\Omega }=2A$

Putting the values for the current $I_{5\Omega }=2A$and $R=5\Omega$into equation (1) to get the dissipated power by resistance $5\Omega$by

$P_{5\Omega }={I^2}_{5\Omega }R={\left(2A\right)}^2\left(5\Omega \right)=20W$

The potential difference in the upper branch could be calculated by

$△V=I(10\Omega +\hspace{0.17em}5\Omega )=\left(2A\right)(10\Omega +5\Omega )=30V$

The potential difference across the resistors is the same in parallel connection. So, the voltage across the below resistor is $30V$. So, this value to get the power through the resistor $20\Omega$by

$P_{20\Omega }=\frac{{\left(△V_R\right)}^2}{R}=\frac{\left(30V\right)}{20\Omega }=45W$