Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentEasy2 minPYQ_2020

PhysicsEasysingle choice

The length of a wire of a potentiometer is $100 cm$ , and the emf of its primary cell is $E volt$ . It is employed to measure the emf of a battery whose internal resistance is $0.5$ $Ω$ . If the balance point is obtained at $l = 30 cm$ from the positive end, the emf of the battery is

Options:

Answer:

Solution:

$V \propto l$

$\Rightarrow \frac{V}{E} = \frac{l}{L}$

Where, $l$ = balance point

L= length of potentiometer wire.

Or $V = \frac{l}{L} E$

$o r V = \frac{30 \times E}{100} = \frac{30}{100} E$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument

⏱ 2mℹ️ Source: PYQ_2020

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