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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical instrumentEasy2 minQB

PhysicsEasysingle choice

A voltmeter with a resistance $50 \times 10^{3} Ω$ is used to measure voltage in a circuit. To increase its range to $3$ times, the additional resistance to be put in series is

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Solution:

$V = i_{g} R a n d V^{'} = i_{g} R^{'} o r \frac{R^{'}}{R} = \frac{V^{'}}{V}$

Or $R^{'} = \frac{V^{'}}{V} R = \frac{3 V}{V} \times 50 \times 10^{3} = 1.5 \times 10^{5} Ω$

$∴$ Additional resistance

$= 1.5 \times 10^{5} - 0.5 \times 10^{5} = 10^{5} Ω$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument

⏱ 2mℹ️ Source: QB

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