Question

A$\mathbf{1}\mathbf{.}\mathbf{58}\mathbf{-}\mathbf{V}$alkaline cell with a$\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{-}\mathbf{\Omega }$internal resistance is supplying$\mathbf{8}\mathbf{.}\mathbf{50}\mathbf{A}$to a load.(a) What is its terminal voltage?(b) What is the value of the load resistance?(c) What is unreasonable about these results?(d) Which assumptions are unreasonable or inconsistent?

Short answer

a) Terminal voltage of the battery is$-0.12\mathrm{V}$.

b) Value of the load resistance is$-0.014\mathrm{\Omega }$.

c) Terminal voltage and resistance are negative which cannot be possible.

d) Maximal current is too high.

Step-by-step solution

Calculation for the terminal voltage of battery.

In this problem we consider a battery with emf $\mathrm{E}=1.58\mathrm{V}$and an internal resistance $\mathrm{r}=0.2\mathrm{\Omega }$which supplies a current $\mathrm{l}=8.5\mathrm{A}$

(a)

The terminal voltage of the battery is obtained as

$$\begin{gathered} \mathrm{V}=\mathrm{E}-\mathrm{lr} \\ =1.58\mathrm{V}-\left(8.5\mathrm{A}\right)\times \left(0.2\mathrm{\Omega }\right) \\ =0.12\mathrm{V} \end{gathered}$$

Calculation for the load resistance.

(b)

The load resistance can be obtained from the loop rule as

$$\begin{gathered} \mathrm{E}=\mathrm{lr}-\mathrm{lR}=0 \\ \mathrm{R}=\frac{\mathrm{E}}{\mathrm{l}}-\mathrm{r} \\ \mathrm{R}=\frac{1.58\mathrm{V}}{8.5\mathrm{A}} \\ \mathrm{R}=0.014\mathrm{\Omega } \end{gathered}$$

Therefore, the load resistance of the battery is $-0.014\Omega$.

Explanation

(c)

Neither of these two results should be negative.

The negative terminal voltage would mean that the effective poles of the battery in comparison to emf, which is impossible. The resistance can never be negative.

Calculation for the maximum current when the external loading is zero.

(d)

The current given in the problem is too large for such a small battery. In the case where the external load is zero, the current for the given emf and internal resistance is

$$\begin{gathered} \mathrm{E}-{\mathrm{l}}_{\max }\mathrm{r}=0 \\ {\mathrm{l}}_{\max }=\frac{\mathrm{E}}{\mathrm{r}} \\ {\mathrm{l}}_{\max }=\frac{1.58\mathrm{V}}{0.2\mathrm{\Omega }} \\ {\mathrm{l}}_{\max }=7.9\mathrm{A} \end{gathered}$$

This is the maximal possible current in the ideal case where there is no external resistance. Even the wires that connect the battery to the load have some resistance so the current should in any case be smaller than this.