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

Option 1 - 2.2W

Option 2 - 0.65W

Option 3 - 4.4W

Option 4 - 1.32W

17 Views|Posted 6 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 1

Detailed Solution:

$R_{e q} = 0.6 Ω + \frac{\frac{4 * 12}{4 + 12} * \frac{6 * 8}{6 + 8}}{\frac{4 * 12}{4 + 12} + \frac{6 * 8}{6 + 8}} Ω$

$= 0.6 Ω + \frac{3 * \frac{48}{14}}{3 + \frac{48}{14}} Ω$

$= 0.6 Ω + \frac{(144 / 14) Ω}{(\frac{90}{14})}$

$\begin{array}{l} = 0.6 Ω + 1.6 Ω \\ = 2.2 Ω \end{array}$

p = $\frac{v^{2}}{R_{e q}} = \frac{{(2.2 V)}^{2}}{2.2 Ω} = 2.2 W$

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

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