Question

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.

Short answer

(a) The amount of current that flows isI=1.58 mA.

(b) The terminal voltage is V=1.5848 V .

(c) The ratio of terminal voltage to the emf is 0.9998 .

Step-by-step solution

Concept Introduction

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.

Calculation for Current

(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 $1.58\mathrm{mA}$.

Calculation for Terminal Voltage

(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 $1.5848\mathrm{V}$.

Calculation for Ratio of Terminal Voltage and EMF

(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 $0.9998$.