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Two resistors 400Omega and 800Omega are ...

Two resistors `400Omega` and `800Omega` are connected in series with a `6V` battery. It is desired to measure the current in the circuit. An ammeter of `10Omega` resistance is used for this purpose figure. What will be the reading in the ammeter? Similarly, if a voltmeter of `10,000Omega` resistance is used to measure the potential difference across the `400Omega` resistor, what will be the reading in the voltmeter?

Text Solution

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Refer figure (a) Current through ammeter
`i=(net emf)/(net resistance)=6/(400+800+10)=4.96 times 10^-3A=4.96mA`
Refer figure (B) Combined resitance of `1000 Omega` voltmeter and `400 Omega` resistance is
`R=(1000 times 400)/(1000+400)=285.71 Omega therefore i=6/((285.71+800))=5.53 times 10^-3 A`
Reading of voltmeter `=V_(ab)=i'R=(5.53 times 10^-3)(285.71)=1.58V`
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