Two resistors `10Omega` and `25Omega` are connected in series and the combination is connected across a 25 V source of negligible internal resistance. A voltmeter of resistance `25Omega` is connected across the `10Omega` resistor. Find the reading in the voltmeter. If another voltmeter of a very large resistance replaces the voltmeter, then find the reading of voltmeter.

Explanation

We know that, `R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))`
i.e., `R_(p)=(25xx10)/35=7.142Omega`
`R_("eq")=7.142+25=32.143Omega`
Current in the circuit `=V/(R_("eq"))=25/(32.143)=0.778A`
p.d across the 25 `Omega` resistor=`0.778xx25=19.44` V
Voltmeter reading =25 - 19.44 = 5.56 V
In the absence of the 25 `Omega` voltmeter,
`I.=25/(10+25)=0.7143 A`