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Q.
The total power dissipated in watt in the circuit shown here is
AIPMTAIPMT 2007Current Electricity
Solution:
The resistance of $6 \, \Omega$ and $3\, \Omega$ are in parallel in the given circuit, their equivalent resistance is
$\frac{1}{R_{1}}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{1}{2}$ or $R_{1}=2 \Omega$
Again, $R_{1}$ is in series with $4 \Omega$ resistance, hence ltbgt
$R=R_{1}+4=2+4=6 \, \Omega$
Thus, the total power dissipatesin the circuit
$P=\frac{V^{2}}{R}$
Here, $V=18 \, V, R=6\, \Omega$
Thus, $P=\frac{(18)^{2}}{6}=54 \, W$
