91Ó°ÊÓ

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

$40W$of the power is dissipated by $10\mathrm{Î\copyright }$resistor. So, calculating the current through the resistance $10\mathrm{Î\copyright }$. 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 $R=10\mathrm{Î\copyright }$and $P_R=40W$to get $I_{10\mathrm{Î\copyright }}$by

$I_{10\mathrm{Î\copyright }}=\sqrt{\frac{P_R}{R}}=\sqrt{\frac{40W}{10\mathrm{Î\copyright }}}=2A$

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

$I_{10\mathrm{Î\copyright }}=I_{5\mathrm{Î\copyright }}=2A$

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

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

The potential difference in the upper branch could be calculated by

$â–³V=I(10\mathrm{Î\copyright }+ 5\mathrm{Î\copyright })=\left(2A\right)(10\mathrm{Î\copyright }+5\mathrm{Î\copyright })=30V$

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 $20\mathrm{Î\copyright }$by

$P_{20\mathrm{Î\copyright }}=\frac{{\left(â–³V_R\right)}^2}{R}=\frac{\left(30V\right)}{20\mathrm{Î\copyright }}=45W$