Two resistors  400Ω and  800Ω are connected in series across a 6V battery. The potential difference measured by a voltmeter of   10kΩ across 400Ω  resistor is close is :

Let voltmeter measure v volts∴i  in 10kΩ  =   i1=v10000Ai  in  400Ω=i2=v400A i in  800Ω=i=i1+i2=v10000+v400AApplying Kirchhoff’s law for loop ABCDEA  v+i800-6=0⇒v+v10000+v400800=6v=600308  v=1.95volts