Question

A car battery with a 12-V emf and an internal resistance of $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}\mathbf{\Omega }$is being charged with a current of 60 A Note that in this process the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted to chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply 60 A to the starter motor?

Short answer

(a) The potential difference across terminals V=15 V .

(b) The rate of the thermal energy ${\mathrm{P}}_{\mathrm{t}}=180\mathrm{W}$.

(c) The rate of the electric energy ${\mathrm{P}}_{\mathrm{c}}=720\mathrm{W}$.

(d) The potential difference and rate of thermal energy are $\mathrm{V}=9\mathrm{V},{\mathrm{P}}_{\mathrm{t}}=180\mathrm{W}$respectively.

Step-by-step solution

Formula of the potential difference

The formula to find the potential difference is,

$\mathbf{∆}\mathbf{V}\mathbf{=}\mathbf{V}\mathbf{+}\mathbf{Ir}$

Here, r is the internal resistance of the battery, I is the current through the battery, and V is the emf.

Potential difference across terminal

The car battery with E=12 V emf and internal resistance$\mathrm{r}=0.05\mathrm{\Omega }$ . The battery is charged with a current of $\mathrm{I}=60\mathrm{A}$.

(a) The battery's terminal voltage is equal to the potential difference across its terminals. Because the current flows in the opposite direction when the battery is charged than when it is discharged, the terminal voltage is higher.

$$\begin{gathered} \mathrm{V}=\mathrm{E}+\mathrm{Ir} \\ =12\mathrm{V}+\left(60\mathrm{A}\right)·\left(0.05\mathrm{\Omega }\right) \\ =15\mathrm{V} \end{gathered}$$

Therefore, the potential difference across terminals V=15 V .

Calculation of the rate of the thermal energy

(b) To charge the battery, the electric energy from the charger must be transformed into chemical energy in the battery. In the resistor, some of this energy is wasted as thermal energy.

The rate of thermal energy dissipation, which is equivalent to the resistance's power, is calculated as follows:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{t}}={\mathrm{I}}^2\mathrm{r} \\ ={\left(60\mathrm{A}\right)}^2·\left(0.05\mathrm{\Omega }\right) \\ =180\mathrm{W} \end{gathered}$$

Therefore, the rate of the thermal energy ${\mathrm{P}}_{\mathrm{t}}=180\mathrm{W}$.

Calculation of the rate of the electrical energy

(c) Assume that the energy from the charger does not leak anywhere, that is, that it is only spent on the charging of the battery and on the internal resistance. First have to calculate the total power, which is the total rate of energy converted from the charger to the battery. This is obtained from the terminal voltage of the battery and the current as

$$\begin{gathered} \mathrm{P}=\mathrm{VI} \\ =\left(15\mathrm{V}\right)·\left(60\mathrm{A}\right) \\ =900\mathrm{W} \end{gathered}$$

This then gives the rate of conversion of electric energy to chemical energy as

$$\begin{gathered} {\mathrm{P}}_{\mathrm{c}}=\mathrm{P}‐{\mathrm{P}}_{\mathrm{t}} \\ =900\mathrm{W}‐180\mathrm{W} \\ =720\mathrm{W} \end{gathered}$$

Therefore the rate of the electric energy ${\mathrm{P}}_{\mathrm{c}}=720\mathrm{W}$.

Calculation of the rate of the thermal energy and electrical energy

(d) When the battery is used to supply a current of I=60 A to the starter motor, the current goes in the usual direction, from the positive pole through the internal resistance and the motor back to the negative pole. The terminal voltage is in this case

$$\begin{gathered} \mathrm{V}=\mathrm{E}+\mathrm{Ir} \\ =12\mathrm{V}‐\left(60\mathrm{A}\right)·\left(0.05\mathrm{\Omega }\right) \\ =9\mathrm{V} \end{gathered}$$

The dissipation rate of the thermal energy is the same as before because the battery supplies the same current as the charger and the internal resistance is the same, which gives

$$\begin{gathered} {\mathrm{P}}_{\mathrm{t}}={\mathrm{I}}^2\mathrm{r} \\ ={\left(60\mathrm{A}\right)}^2·\left(0.05\mathrm{\Omega }\right) \\ =180\mathrm{W} \end{gathered}$$

Therefore the potential difference and rate of thermal energy are $\mathrm{V}=9\mathrm{V},{\mathrm{P}}_{\mathrm{t}}=180\mathrm{W}$respectively.