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PhysicsCurrent ElectricityElectrical instrumentMedium2 minPYQ_2024
PhysicsMediumnumerical

Two resistance of100Ωand200Ωare connected in series with a battery of4 Vand negligible internal resistance. A voltmeter is used to measure voltage across100Ωresistance, which gives reading as1 V. The resistance of voltmeter must be _______Ω.

Question diagram: Two resistance of 100 Ω and 200 Ω are connected in series wi
Answer:
200.00
Solution:

Voltage across 200 Ω=4-1=3 V.

Therefore, current through the battery, 3200

Now, equivalent resistance of RV & 100 Ω,=Rv100Rv+100

For voltmeter, Rv100Rv+100×3200=1

3Rv=2Rv+200

Rv=200 Ω

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical instrument
2mℹ️ Source: PYQ_2024

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