. The resistance of the coil of a pivotedcoil galvanometer is${\mathbf{R}}_{\mathbf{c}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{36}\mathbf{\Omega }$, and a current of $\mathbf{fs}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0224}\mathbf{A}$causes it to deflect full scale. We want to convert this galvanometer to an ammeter readingn $\mathbf{a}\mathbf{=}\mathbf{20}\mathbf{A}$full scale. The only shunt available has a resistance of ${\mathbf{R}}_{\mathbf{sh}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{\Omega }$. What resistance Rmust be connected in series with the coil (Fig. E26.36)?

A galvanometer is a device that is used to detect small electric current or measure its magnitude.The current and its intensity is usually indicated by a magnetic needle’s movement or that of a coil in a magnetic field that is an important part of a galvanometer.

Given
We are given the resistance of the ${\mathrm{R}}_{\mathrm{c}}=9.36\mathrm{\Omega }$coil and the fullscale deflection current is ${\mathrm{l}}_{\mathrm{fs}}=0.0224\mathrm{A}$and themaximum current required to read is and the shunt resistance is ${\mathrm{R}}_{\mathrm{sh}}=0.025\mathrm{\Omega }$

Solution

We Want to determine the series resistance R in the figure. R is in series With Rc and at the left black point on the figure, anexternal current enters at this junction point, where a part of the current goes through R and a part goes to R31, so We apply the junction rule in this case, where the junction rule is based on conservation of electric charge and the current entersajunction point is equal to the current ?oWs out from this point, so in our circuit, We could get the current ?ow in R31], whereIa enters the junction point as shown in the figure While flow out from it
${\mathrm{l}}_{\mathrm{a}}={\mathrm{l}}_{\mathrm{r}}+{\mathrm{l}}_{\mathrm{sh}}$
As R and Rc are in series, therefore they have the same current and , so We can solve this equation for I311 and plug
our values for ${\mathrm{l}}_{\mathrm{sh}}={\mathrm{l}}_{\mathrm{a}}+{\mathrm{l}}_{\mathrm{R}}$
$$\begin{gathered} {\mathrm{l}}_{\mathrm{sh}}={\mathrm{l}}_{\mathrm{a}}-{\mathrm{l}}_{\mathrm{R}} \\ =20\mathrm{A}-0.224\mathrm{A} \\ =19.98\mathrm{A} \end{gathered}$$
Calculating the current in the shunt resistance will find the voltage across it by using Ohm's law in the next
$$\begin{gathered} {\mathrm{V}}_{\mathrm{sh}}={\mathrm{l}}_{\mathrm{sh}}{\mathrm{R}}_{\mathrm{sh}} \\ =19.98\left(0.025\right) \\ =0.4995\mathrm{V} \end{gathered}$$

theshuntresistanceisinparallelwiththecombinationofbothresistorsRandR0therefore,thecombinationhasthesamevoltageofR3,asnext
${\mathrm{V}}_{\mathrm{sh}}={\mathrm{I}}_{\mathrm{fs}}({\mathrm{R}}_{\mathrm{c}}+\mathrm{R})$
Now We can plug our values for , Vsh and Rc into equation (1) to get R
$\mathrm{R}=\frac{{\mathrm{V}}_{\mathrm{sh}}}{{\mathrm{I}}_{\mathrm{fs}}}-{\mathrm{R}}_{\mathrm{c}}$
Now we can plug our values for Ifs, V311 and Rc into equation (1) to get R
$$\begin{gathered} \mathrm{R}=\frac{{\mathrm{V}}_{\mathrm{sh}}}{{\mathrm{I}}_{\mathrm{fs}}}-{\mathrm{R}}_{\mathrm{c}} \\ -=\frac{0.4995}{0.0224}-9.36 \\ =12.94\mathrm{\Omega } \end{gathered}$$