Question

"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0% of the emf of the battery. Express your answer numerically (in ohms) to at least three significant digits.

Answer 1

The minimum value is
`R_V =44.552\ \Omega`

Step-by-step explanation:

From the question we are given that

The voltage is
`E = 7.5V`

The internal resistance is
`r = 0.45`

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

What is means is that we need to obtain voltmeter resistance such that

V = (100% -1%) of E

Where E is the e.m.f of the battery and V is the voltmeter reading

i.e V = 99% of E = 0.99 E = 7.425

Generally

E = V + ir

where ir is the internal potential difference of the voltmeter and

V is the voltmeter reading

Making i the subject of the formula above

`i = ((E-V))/(r)`

`=(7.50-7.425)/(0.45)`

`= 0.1667 A`

Now the current is constant through out the circuit so,

`V = iR_V`

Where
`R_V` is the value of voltmeter resistance

Hence
`R_V = (V)/(i) = (7.425)/(0.1667)`

`=44.552\ \Omega`