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PhysicsCurrent ElectricityEquivalent ResistanceEasy2 minPYQ_2020

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A $6 V$ battery is connected to the terminals of the three-metre long wire of uniform thickness and resistance of $100 Ω$ . The difference of potential between two points on the wire separated by a distance of $50 cm$ will be

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Answer:

Solution:

$R \propto ℓ$

For $300 cm$ , $R = 100 Ω$

For $50 cm$ , $R^{'} = \frac{100}{300} \times 50 = \frac{50}{3} Ω$

$∴ IR = 6$

$\Rightarrow {IR}^{'} = \frac{6}{R} \times R^{'} = \frac{6}{100} \times \frac{50}{3} = 1 V$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Equivalent Resistance

⏱ 2mℹ️ Source: PYQ_2020

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