A$\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{}\mathbf{\Omega }$ammeter is placed in series with a$\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{\Omega }$resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) Calculate the resistance of the combination. (c) If the voltage is kept the same across the combination as it was through the$\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{\Omega }$resistor alone, what is the percent decrease in current? (d) If the current is kept the same through the combination as it was through the$\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{‐}\mathbf{\Omega }$resistor alone, what is the percent increase in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss.

The rate of electron flow can be described as the aggregate flow of electrons via a wire. The term "resistance" refers to anything that stands in the way of current flow. To convert electrical energy to light, heat, or movement, an electrical circuit must have resistance.

(a)

The circuit diagram of the connection of an ammeter when a $0.02\mathrm{\Omega }$ammeter is placed in series with a$10\mathrm{\Omega }$resistor in a circuit is given below,

Therefore, the circuit diagram is obtained.

(b)

A $0.02\mathrm{\Omega }$ammeter is placed in series with a $10\mathrm{\Omega }$resistor in a circuit.

The formula for total resistance is,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{combination}}=\mathrm{R}+\mathrm{r} \\ =10\mathrm{\Omega }+0.02\mathrm{\Omega } \\ =10.02\mathrm{\Omega } \end{gathered}$$

Therefore, the value for total resistance is ${\mathrm{R}}_{\mathrm{combination}}=10.02\mathrm{\Omega }$.

(c)

Current through the only resistor is,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ =\frac{\mathrm{V}}{10\mathrm{\Omega }} \end{gathered}$$

And current through combination is,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{combination}}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{combination}}} \\ =\frac{\mathrm{V}}{10.02\mathrm{\Omega }} \end{gathered}$$

The decrease in current is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{d}}=\left(\frac{{\mathrm{I}}_{\mathrm{combination}}-\mathrm{I}}{\mathrm{I}}\right)\times 100\% \\ =\left(\frac{\left({\displaystyle \frac{\mathrm{V}}{10.02\mathrm{\Omega }}}\right)-\left({\displaystyle \frac{\mathrm{V}}{10\mathrm{\Omega }}}\right)}{\left({\displaystyle \frac{\mathrm{V}}{10\mathrm{\Omega }}}\right)}\right)\times 100\% \\ =-0.1996\% \end{gathered}$$

Therefore, the decrease in current is $0.1996\%$.

(d)

Voltage through the only resistor is,

$$\begin{gathered} \mathrm{V}=\mathrm{IR} \\ =\mathrm{I}\times 10\mathrm{\Omega } \end{gathered}$$

And voltage through combination is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{Combination}}={\mathrm{IR}}_{\mathrm{Combination}} \\ =\mathrm{I}\times 10.02\mathrm{\Omega } \end{gathered}$$

The increase in voltage is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{I}}=\left(\frac{{\mathrm{V}}_{\mathrm{Combination}}-\mathrm{V}}{\mathrm{V}}\right)\times 100\% \\ =\left(\frac{\left(\mathrm{I}\times 10.02\mathrm{\Omega }\right)-\left(\mathrm{I}\times 10\mathrm{\Omega }\right)}{\left(\mathrm{I}\times 10\mathrm{\Omega }\right)}\right)\times 100\% \\ =0.2\% \end{gathered}$$

Therefore, the increase in voltage is $0.2\%$.

(e)

The result for the decrease in current is $0.1996\%$.

The result ofthe increase in voltage is $0.2\%$.

These are very small amounts, so they do not affect the circuit in anyway.

Therefore, the results are not significant.