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Question
Figure below shows two resistors R1 and R2 connected to a battery having an emf of 40V and negligible internal resistance. A voltmeter having a resistance of. 300 Ω is used to measure the potential difference across R1 Find the reading of the voltmeter.
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Solution
R1 and R3 are in parallel.
∴ Current I in the circuit remains unchanged in R1 and R3 combination and R2
`1/(R') = 1/200 +1/300 =5/600`
`R' = 600/5 = 120 Omega `
∴ REquivalent in circuit = 120 + 880 = 1000 Ω
I, current.in the circuit =`40/1000 A`
∴ V =` 40/1000 xx 120`
⇒ `V = 48/10`
∴ V = 4.8 V
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