Question

The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes10.20 V . (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short answer

1. The internal resistance of the battery is$0.36\Omega$.
2. The emf of the battery is 8.94 V .

Step-by-step solution

Define the ohm’s Law.

Consider the real source of emf has internal energy then its terminal potential difference$V_{ab}$ is as follows:

${\mathit{V}}_{\mathbf{a}\mathbf{b}}\mathbf{=}\mathit{\epsilon }\mathbf{-}\mathbf{IR}$ ….. (1)

Determine the internal resistance.

(a)

Consider the given expression for the current and the voltage.

$$\begin{gathered} V_{ba}=8.40V \\ {I}_{ba}=1.5A \end{gathered}$$

Substitute the values in the equation (1).

$$\begin{gathered} 8.40=\epsilon -(1.5)\left(R\right) \\ \epsilon =8.40+1.5r \end{gathered}$$

Consider the current is 3.50 A in the reverse direction, the potential difference becomes 10.20 V.

Substitute the values in the equation $V=\epsilon +lr$and solve.

$$\begin{gathered} 10.20=(8.30+1.5\mathrm{r})+3.5\mathrm{r} \\ 5\mathrm{r}=1.8 \\ \mathrm{r}=0.36\mathrm{\Omega } \end{gathered}$$

Hence, the internal resistance of the battery is$0.36\mathrm{\Omega }$

Determine the emf of battery.

(b)

Given: the value of the internal resistance is $0.36\mathrm{\Omega }$.

Substitute $0.36\mathrm{\Omega }$for r in the equation $\epsilon =8.40+1.5r$.

$$\begin{gathered} \epsilon =8.40+1.5\times 0.36 \\ =8.94V \end{gathered}$$

Hence, the emf of the battery is 8.94 V.