Question: In Fig. 27-14, assume that$\mathit{\epsilon }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{,}\mathit{r}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{,}{\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{,}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{\Omega }$If the ammeter resistance R_Ais $\mathbf{}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{\Omega }$, what percent error does it introduce into the measurement of the current? Assume that the voltmeter is not present.

1. The given resistances are: , $\mathbf{}\mathit{r}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{,}{\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{,}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{\Omega }$,.
2. The emf of the battery,$\mathit{\epsilon }\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}$ .
3. The resistance of the ammeter, ${\mathit{R}}_{\mathbf{A}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathit{\Omega }$.

The threshold voltage where the breakdown of the circuit initiates is called the breakdown voltage of the circuit. The minimum voltage at which an insulator experiences momentary conduction is the breakdown voltage. Using the potential difference concept, we can get the resistance of the resistor in the given circuit.

Formulae:

The voltage equation using Ohm’s law,

V =IR

(i)

Here I is the current, and R is the resistance.

The error in of a value,

$\text{Error}\%=\frac{\text{Initial}\text{value}-\text{New}\text{value}}{\text{Initial}\text{value}}\times 100\%$

Now, the current value without ammeter resistance can be given using equation (i) as follows:

$i=\frac{\epsilon }{r+R_1+R_2}$

(a)

Now, the current value with ammeter resistance can be given using equation (i) as follows:

$i_A=\frac{\epsilon }{r+R_1+R_2+R_A}$ (b)

Thus, the error percentage can be given using equations (a) and (b) in equation (ii) as follows:

$$\begin{gathered} \text{Error}\%=\frac{i_A-i}{i}\times 100\% \\ =\frac{\left(\frac{\epsilon }{r+R_1+R_2}-\frac{\epsilon }{r+R_1+R_2+R_A}\right)}{\frac{\epsilon }{r+R_1+R_2}}\times 100\% \\ =\left(1-\frac{r+R_1+R_2}{r+R_1+R_2+R_A}\right)\times 100\% \end{gathered}$$

Substitute the values in the above formula, and we get,

$$\begin{gathered} \text{Error}\%=\frac{R_A}{r+R_1+R_2+R_A}\times 100\% \\ =\frac{0.10\Omega }{\text{2.0}\Omega +5.0\Omega +\text{4.0}\Omega +0.10\Omega }\times 100\% \\ =0.901\% \\ \simeq 0.90\% \end{gathered}$$

Hence, the value of the error percentage in the current is 0.90%.