A \({\rm{1}}{\rm{.00}}\;{\rm{M\Omega }}\) voltmeter is placed in parallel with a \({\rm{75}}{\rm{.0}}\;{\rm{k\Omega }}\) resistor in a circuit. (a) Draw a circuit diagram of the connection.

(b) What is the resistance of the combination?

(c) If the voltage across the combination is kept the same as it was across the \({\rm{75}}{\rm{.0}}\;{\rm{k\Omega }}\) resistor alone, what is the percent increase in current?

(d) If the current through the combination is kept the same as it was through the \({\rm{75}}{\rm{.0}}\;{\rm{k\Omega }}\) resistor alone, what is the percentage decrease in voltage?

(e) Are the changes found in parts (c) and (d) significant? Discuss.

Internal resistance of the voltmeter\(r = 1\;{\rm{M}}\Omega \)

Load resistance is \(R = 75\;{\rm{k}}\Omega \)

If a high value resistance is added in series with a galvanometer, the setup can be used in place of a voltmeter. The higher the sensitivity of the galvanometer, the lower value of potential difference it can measure and the lower the resistivity, only higher values it can measure.

(a)

The circuit diagram of the connection of voltmeter is given below –

(b)

The formula for total resistance of the combination is –

\(\begin{array}{c}{R_{comb }} = \frac{{Rr}}{{R + r}}\\ = \frac{{{{10}^6}\;\Omega \times 75 \times {{10}^3}\;\Omega }}{{{{10}^6}\;\Omega + 75 \times {{10}^3}\;\Omega }}\\ = 69797\;\Omega \end{array}\)

Therefore, the value for total resistance is\({R_{comb }} = 69797\;\Omega \).

(c)

If the voltage across the circuit is kept the same as when only load resistance was present, then current in circuit will be-

\(\begin{array}{c}I = \frac{{\;V}}{R}\\ = \frac{V}{{75 \times {{10}^3}\;\Omega }}\end{array}\)

And current through combination is –

\({I_{comb }} = \frac{V}{{{R_{comb }}}} = \frac{V}{{69797\;\Omega }}\)

The increase in current is –

\(\begin{array}{c}\% \;in{c_i} = \frac{{{I_{comb}} - I}}{I} \times 100\\ = \frac{{\frac{V}{{69797\;\Omega }} - \frac{V}{{75 \times {{10}^3}\;\Omega }}}}{{\frac{V}{{75 \times {{10}^3}\;\Omega }}}} \times 100\\ = 7.45\;\% \end{array}\)

Therefore, the percentage increase in current is \(\% \;in{c_i} = 7.45\;\% \).

(d)

If the current across the circuit is kept the same as when only load resistance was present, then voltage in circuit will be-

\(\begin{array}{c}V = IR\\ = \left( {75 \times {{10}^3}\;\Omega } \right)I\end{array}\)

voltage through combination is –

\(\begin{array}{c}{V_{comb}} = I{R_{comb }}\\ = \left( {69797\;\Omega } \right)I\end{array}\)

The decrease in voltage is –

\(\begin{array}{c}\% \;de{c_V} = \frac{{V - {V_{comb}}}}{{{V_{comb}}}}\\ = \frac{{\left( {75 \times {{10}^3}\;\Omega } \right)I - \left( {69797\;\Omega } \right)I}}{{\left( {75 \times {{10}^3}\;\Omega } \right)I}}\\ = 6.94\;\% \end{array}\)

Therefore, the percentage decrease in voltage is \(\% \;de{c_V} = 6.94\;\% \).

(e)

The result for increase in current is\(\% \;in{c_i} = 7.45\;\% \).

The result for decrease in voltage is\(\% \;de{c_V} = 6.94\;\% \).

The values are significant because the current consumption increases when the voltmeter takes the reading, this introduces an error in the reading.