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: (a) 1.32 W (b) 0.65 W (c) 2.2 W (d) 4.4 W

Answer: (c). AB4 Ω8 Ω12 Ω6 ΩR = 1.6 Ω The left side is node A and the right side is node B, so every branch runs from A to B in parallel. Branches: top 4 Ω, bottom 8 Ω, crossed 4 + 8 = 12 Ω and 4 + 2 = 6 Ω. 1/R = 14 + 18 + 112 + 16 = 15/24 ⇒ R = 1.6 Ω. I = (2.2)/(1.6 + 0.6) = 1 A, total power P = EI = 2.2 W. So the answer is (c) 2.2 W.

In the given figure, the emf of the cell is 2.2 V and if internal… | JEE Main 2021 (26 Aug, Shift 1)

Q33JEE Main 2021 · 26 AUG · SHIFT 1Medium

Electrical Power & Heating Effect

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

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Answer: (c)

The left side is node A and the right side is node B, so every branch runs from A to B in parallel.
Branches: top $4\ \Omega$ , bottom $8\ \Omega$ , crossed $4 + 8 = 12\ \Omega$ and $4 + 2 = 6\ \Omega$ .
$\dfrac{1}{R} = \dfrac14 + \dfrac18 + \dfrac1{12} + \dfrac16 = \dfrac{15}{24} \Rightarrow R = 1.6\ \Omega$ .
$I = \dfrac{2.2}{1.6 + 0.6} = 1\text{ A}$ , total power $P = EI = 2.2\text{ W}$ .
So the answer is (c) 2.2 W.

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Electrical Power & Heating Effect question from JEE Main 2021 (26 Aug, Shift 1)

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