Question

A battery of emf `12V` and internal resistance `2 Omega` is connected two a `4 Omega` resistor. Show that the a voltmeter when placed across cell and across the resistor in turn given the same reading

Answer 1

Total resistance `=2+4=6Omega`
`(a) 1=(E)/(R )=(12)/(6)=2A`
`V=E-Ir=12-2xx2=*"Volt"`
`"Voltmeter reading"=8V`
`"Ammeter reading"=2A`
Now when voltmeter is put parallel to `4Omega` resistance, potential drop across
`4Omega` resistance is `4xx2=8V`. So, voltmeter will show `8V`.
`(b)` Voltmeter is placed parallel and ammeter in series in the circuit because ammeter is a very low resistance device and voltmeter has very high resistance.