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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityBatteryEasy2 minQB

PhysicsEasysingle choice

A cell $E_{1}$ of emf $6 V$ and internal resistance $2Ω$ is connected with another cell $E_{2}$ of emf $4 V$ and internal resistance $8Ω$ (as shown in the figure). The potential difference across points $X$ and $Y$ is:

Options:

Answer:

Solution:

$I = \frac{6 - 4}{10} = \frac{1}{5} A$

$V_{x} + 4 + 8 \times \frac{1}{5} - V_{y} = 0$

$V_{x} - V_{y} = - 5.6 \Rightarrow ∣ V_{x} - V_{y} ∣ = 5.6 V$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Battery

⏱ 2mℹ️ Source: QB

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