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A voltmeter resistance 500Omega is used to measure the emf of a cell of internal resistance 4Omega. The percentage error in the reading of the voltmeter will be |
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Answer» Solution :`V=E-ir` `THEREFORE"PERCENTAGE ERROR "=(DeltaE)/(E)xx100=(ir)/(E)xx100` `=((E)/(R+r)r)/(E)xx100=((r)/(R+r))xx100` `=((4)/(500+4))xx100=0.8%` |
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