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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityCircuit AnalysisEasy2 minPYQ_2020

PhysicsEasysingle choice

In the following network, the potential at $O$ is

Options:

Answer:

Solution:

Let the potential at O is $V_{0} .$

Application of Kirchhoff's first law at junction O gives

$\frac{8 - V_{0}}{2} = \frac{V_{0} - 4}{4} + \frac{V_{0} - 2}{2}$

$= \frac{V_{0} - 4 + 2 V_{0} - 4}{4}$

$\frac{4 (8 - V_{0})}{2} = V_{0} - 4 + 2 V_{0} - 4$

$16 - 2 V_{0} = 3 V_{0} - 8$

$16 + 8 = 5 V_{0}$

$V_{0} = \frac{24}{5} = 4 .8 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Circuit Analysis

⏱ 2mℹ️ Source: PYQ_2020

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