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Q.
In the circuit shown, the cell of emf $6V$ is ideal. The resistor in which the power dissipated is greatest, is
NTA AbhyasNTA Abhyas 2020
Solution:
The circuit can be drawn as
As $6\Omega$ an $3\Omega$ are in parallel
As $18\Omega$ and $9\Omega$ are in parallel. Also both $2\Omega$ are in parallel
At point $P$ potential is equally divided as resistance of both parts are same $\left(\right.6\Omega\left.\right)$ .
Equivalent resistance ( $R_{e q}$ ) of the circuit is $12\Omega$ .
Current through the battery $I=\frac{V}{R_{e q}}=\frac{6}{12}=\frac{1}{2}A$
So power in $5\Omega$ is maximum
