Question

A galvanometer having a resistance of ${\mathbf{R}}_{\mathbf{c}}\mathbf{=}\mathbf{25}\mathbf{\Omega }$has a ${\mathbf{R}}_{\mathbf{sh}}\mathbf{=}\mathbf{1}\mathbf{\Omega }$shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a $\hat{\mathbf{i}}\mathbf{=}\mathbf{25}\mathbf{V}$circuit consisting of a resistor connected across the terminals of a $\mathbf{R}\mathbf{=}\mathbf{15}\mathbf{\Omega }$battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the truecurrent in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the truecurrent?

Short answer

the current is 1.56A measured by the ammeter.

the true current in the circuit is 1.66A

the percentage error is 6%

Step-by-step solution

About Galvanometer

Galvanometer, instrument for measuring a small electrical current or a function of the current by deflection of a moving coil.The deflection is a mechanical rotation derived from forces resulting from the current.

Step 2 : Determine the Current measure by the ammeter

Given
We are given the resistance of the galvanometer${\mathrm{R}}_{\mathrm{c}}=25\mathrm{\Omega }$and a shunt resistance${\mathrm{R}}_{\mathrm{sh}}=1\mathrm{\Omega }$ This ammeter measure
the current in a circuit$\hat{\mathrm{I}}=25\mathrm{V}$with and$\mathrm{R}=15\mathrm{\Omega }$
Solution
(a) We want to measure the current I measured by the ammeter. The shunt resistance is connected in parallel with R toconvert the galvanometer into an ammeter, where the equivalent resistance of two resistors in parallel is given by equation
26.5 in the next form
$$\begin{gathered} {\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{R}}_{\mathrm{sh}}{\mathrm{R}}_{\mathrm{c}}}{{\mathrm{R}}_{\mathrm{sh}}+{\mathrm{R}}_{\mathrm{c}}} \\ =\frac{1\mathrm{K\Omega }\times 25\mathrm{K\Omega }}{1+25} \\ =0.916\mathrm{K\Omega } \end{gathered}$$

Theammeterwillreadthecurrentduetothiscombinationofresistance-ThecurrentflowsinRisthesameforthecombinedresistance0.961kohmsasbothresistorsareinseriesandwecanmeasurethiscurrentusingOhm'slawinthenextform

Therefore the current is1.56A measured by the ammeter.

Step 3: Determine the true current in thye circuit

Theammeterwillreadthecurrentduetothiscombinationofresistance-ThecurrentflowsinRisthesameforthecombinedresistance0.961kohmsasbothresistorsareinseriesandwecanmeasurethiscurrentusingOhm'slawinthenextform
$$\begin{gathered} \mathrm{I}=\frac{\hat{\mathrm{I}}}{\mathrm{R}+{\mathrm{R}}_{\mathrm{eq}}} \\ =\frac{25}{15+0.916} \\ =1.56\mathrm{A} \end{gathered}$$
Therefore the current is1.56A measured by the ammeter.
Step 3: Determine the true current in thye circuit

(b)ThetruecurrentisthecurrentduetoonlytheresistanceRintheabsenceoftheammeter(RcandRshareabsence)-
We can use Ohm's law again to measure this current in the next form

$$\begin{gathered} {\mathrm{R}}_{\mathrm{c}},{\mathrm{R}}_{\mathrm{sh}} \\ \mathrm{I}=\frac{\hat{\mathrm{I}}}{\mathrm{R}} \\ =\frac{25}{6+15} \\ =1.66\mathrm{A} \\ ∆={\mathrm{I}}_{\mathrm{true}}-{\mathrm{I}}_{\mathrm{meter}} \\ =1.66\mathrm{A}-1.56\mathrm{A} \\ =0.10\mathrm{A} \end{gathered}$$

Therefore the true current in the circuit is 1.66 A
Step 4:Determine the percentage error
(c) There is an error in the reading of current due to the resistance of meter. The difference betWeen the current measured by
the meter and the true current is
The error represents the ratio betWeen this difference and the true current
$$\begin{gathered} \%\mathrm{error}=\frac{∆}{1.66\mathrm{A}} \\ =6\% \end{gathered}$$
Therefore the percentage error is 6%