Two resistors `400Omega` and `800Omega` are connected in series with a 6 V battery. The potential difference measured by voltmeter of `10kOmega` across 400`Omega` resistor is

Explanation

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`