Q 11 :    

Two resistance of 100Ω and 200Ω are connected in series with a battery of 4V and negligible internal resistance. A voltmeter is used to measure voltage across 100Ω resistance, which gives reading as 1V. The resistance of voltmeter must be ________ Ω.             [2024]



(200)

Given potential across Voltmeter = 1V

Current in 100Ω=1100A=i1

200Ω=3100Aas total potential =4V1&1V is drop across 100Ω

Let current in voltmeter = i

i1+i2=i0

i2+1100=3200

i2+3200-1100=1200A1

V=IR11=1200(r),r=200Ω