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

Question

In the given figure, the emf of the cell is 2 . 2 V and if internal resistance is 0 . 6 Ω . Calculate the power dissipated in the whole circuit:

TestHub · Accepted Answer

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,

$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} + \frac{1}{8} \Rightarrow R_{P} = \frac{48}{30} Ω$

Now the net resistance will be, $R_{n e t} = (\frac{48}{30} + 0.6) Ω$ , so the net power, $P = \frac{V^{2}}{R} = \frac{(2.2)^{2}}{2.2} = 2.2 W$

Physics - Current Electricity Question with Solution | TestHub

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