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

Answer:

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

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

            $i = \frac{(E-V)}{r}$

               $=\frac{7.50-7.425}{0.45}$

              $= 0.1667 A$

Now the current is constant through out the circuit so,

                  $V = iR_V$

Where  $R_V$ is the value of voltmeter resistance

                Hence $R_V = \frac{V}{i} = \frac{7.425}{0.1667}$

                                  $=44.552\ \Omega$