Question

A current of $3$A flows through the $2\Omega$ resistor shown in the circuit. The power resistant is the $5\Omega$ resistor is?

Answer 1

Since the 1$\Omega$ and 5$\Omega$ resistors are in series. Current flow will be same in that branch.

Let supply voltage be V and current in 3rd branch is $I_3$.

$I_3$= $\frac{V}{(1+5)\Omega }=\frac{V}{6\Omega }$

$I_3$=$\frac{1}{6}A$

Power at 5$\Omega$ is, P= $I_3^{2}R$

P= ${\left(\frac{1}{6}\right)}^2$ $5\Omega$

=$\frac{5}{36}$ Watt