Two resistances of 400 and 800 are connected in series with 6 ~V battery of negligible internal resistance. A voltmeter of resistance 10000 is used to measure the potential difference across 400 . The error in the measurement of potential difference in volts approximately is :

Step-by-step solution

R₁=400 , R₂=800 P D across 400 resistance (when voltmeter is not connected) V₁= 6 (400+800) 400= 6 400 1200 =2 ~V when voltmeter is connected Total resistance of the circuit, R= ( 10000 400 10000+400 )+800= 10000 400 10400 +800= 40000 104 +800= 40000+83200 104 = 123200 104 = 30800 26 = 15400 13 New P D across 400 resistance, V₂=6-800 ( 6 15400 / 13 )=6- 800 13 6 15400 =6-4.052=1.95 ~V Error =V₁-V₂=2-1.95=0.05