Question 138 of 380

In the given figure, the emf of the cell is $2.2V$ and if internal resistance is $0.6\Omega$. Calculate the power dissipated in the whole circuit:

A$1.32W$B$4.4W$C$0.65W$D$2.2W$

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Answer: (D) $2.2W$

The above circuit in the question is arranged like that, so all the resistance are connected in parallel and the net resistance of the parallel resistor becomes,

$\dfrac{1}{{R}_{P}}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}+\dfrac{1}{{R}_{3}}+\dfrac{1}{{R}_{4}}=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{8} \Rightarrow {R}_{P}=\dfrac{48}{30}\Omega$

Now the net resistance will be, ${R}_{net}=\left(\dfrac{48}{30}+0.6\right)\Omega$, so the net power, $P=\dfrac{{V}^{2}}{R}=\dfrac{(2.2{)}^{2}}{2.2}=2.2W$

JEE Main Physics practice question on Current Electricity (Combination of Resistances), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.