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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityBatteryEasy2 minQB

PhysicsEasysingle choice

In the given circuit, an ideal voltmeter connected across the $10Ω$ resistance reads $2 V$ . The internal resistance $r$ , of each cell is:

Options:

Answer:

Solution:

Current through battery $= \frac{2}{10} + \frac{2}{15} = \frac{6 + 4}{30} = \frac{1}{3}$ A
Also, net resistance $R_{e q} = \frac{15 \times 10}{15 + 10} + 2 + 2 r$
$= 8 + 2 r$
$\Rightarrow \frac{1}{3} = \frac{1.5 + 1.5}{8 + 2 r}$
$8 + 2 r = 9$
$r = \frac{1}{2} = 0.5Ω$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Battery

⏱ 2mℹ️ Source: QB

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