(a) A voltmeter with resistance `R_v` is connected across the terminals of a battery of emf `E` and internal resistance `r`. Find the potential difference measured by the voltmeter.
(b) If` E = 7.50 V` and `r = 0.45 Omega`, find the minimum value of the voltmeter resistance `R_v` so that the voltmeter reading is within `1.0%` of the emf of the battery. (c) Explain why your answer in part (b) represents a minimum value.

Answer

A, B, C, D

Explanation

a. `V=E-ir`
`=E-(E/(r+R_V))r`
`V=((ER_V)/(r+R_V))`………….i
b. `V=E/100`
substituting in eq. i we get
`R_V=4.5xx10^-3Omega`
c. `V=E(1/(1+r/R_V))`
If `R_V` is increased from this value `V` will increased