Suppose you measure the terminal voltage of an1.585-V alkaline cell having an internal resistance of$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{\Omega }$by placing arole="math" localid="1656394302991" $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{k}\mathbf{}\mathbf{\Omega }$voltmeter across its terminals. (See Figure 21.54.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.

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)

Emf for alkaline cell: $\mathrm{E}=1.585\mathrm{V}$

Internal Resistance of the cell: $\mathrm{R}=0.1\mathrm{\Omega }$

Resistance of the voltmeter: $\mathrm{R}=1\mathrm{k\Omega }\left(\frac{{10}^3\mathrm{\Omega }}{1\mathrm{k\Omega }}\right)=1\times {10}^3\mathrm{\Omega }$

To calculate the current that flows through the circuit, use the loop rule to obtain such that,

$$\begin{gathered} \mathrm{E}‐\mathrm{Ir}‐\mathrm{IR}=0 \\ \mathrm{I}=\frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}} \\ \mathrm{I}=\frac{1.585\mathrm{V}}{\left(1\times {10}^3\mathrm{\Omega }+0.1\mathrm{\Omega }\right)} \\ \mathrm{I}=1.58\times {10}^{-3}\mathrm{A}\left(\frac{1\mathrm{mA}}{{10}^{-3}\mathrm{A}}\right) \\ \mathrm{I}=1.58\mathrm{mA} \end{gathered}$$

Therefore, the value for current is obtained as $1.58\mathrm{mA}$.

(b)

The terminal voltage is obtained as,

$$\begin{gathered} \mathrm{V}=\mathrm{E}‐\mathrm{Ir} \\ =1.585\mathrm{V}-\left(1.58\times {10}^{‐3}\mathrm{A}\right)\times \left(0.1\mathrm{\Omega }\right) \\ =1.5848\mathrm{V} \end{gathered}$$

Therefore, the value for terminal voltage is obtained as $1.5848\mathrm{V}$.

(c)

The difference between the terminal voltage and emf is very small due to the small current and internal resistance. The ratio of the terminal voltage and the emf is,

$$\begin{gathered} \frac{\mathrm{V}}{\mathrm{E}}=\frac{1.5848\mathrm{V}}{1.585} \\ =0.9998 \end{gathered}$$

Therefore, the value for the ratio is obtained as $0.9998$.