Question

A car battery with a 12 V emf and an internal resistance of 0.040 Ω is being charged with a current of 50A. What are (a) The potential difference Vacross the terminals, (b) The rate P_fof energy dissipation inside the battery, and(c) The rate P_emfof energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) Vand (e) P_r?

Short answer

1. The potential difference Vacross the terminals is V=14V
2. The rate P_rof energy dissipation inside the battery is$1.0\times {10}^{2}W$
3. The rate of the energy conversion to the chemical form is$6\times {10}^{2}W$
4. The potential difference between the terminals is 10V
5. The rate energy dissipation is$P_r=1.0\times {10}^{2}W.$

Step-by-step solution

Given

Emf$\epsilon =12V$

Internal resistancerole="math" localid="1662562681995" $r=0.040\Omega$

Current i=50A

Determining the concept

The formula for potential difference for a non-ideal battery to calculate the potential difference across the terminal of the battery. Using the power relation, find the energy dissipation inside the battery and the rate of energy conversion to the chemical form.

Formulae are as follow:

$$\begin{gathered} V=\epsilon -ir \\ P=i^{2}r \\ P=iV \end{gathered}$$

Where, P is power, I is current, R is resistance, 𝛆 is emf, V is voltage .

(a) Determining the potential difference across the terminals

The potential difference across the terminals:

The potential difference between the terminals is given by,

$$\begin{gathered} V=\epsilon +ir \\ V=12V+\left(50A\right)(0.040\Omega ) \\ V=14V \end{gathered}$$

Hence, the potential difference Vacross the terminals is 14V.

(b) determining the rate Pr of energy dissipation inside the battery

The rate of energy dissipation inside the battery

The rate energy dissipation is given by,

$P_r=i^{2}r=\left(50A{)}^2\right(0.040\Omega )=1.0\times {10}^{2}W$

Hence, the rate P_rof energy dissipation inside the battery is$1.0\times {10}^{2}W$

(c) Determining the rate Pemf of the energy conversion to the chemical form

The rate P_emfof the energy conversion to the chemical form :

$$\begin{gathered} P=iV \\ P=50A\times 12V \\ P=600W=6\times {10}^{2}W \end{gathered}$$

Hence, the rate P_emfof the energy conversion to the chemical form is$6\times {10}^{2}W$.

(d) determining the potential difference

The current is 50A, then potential difference is as follows:

The potential difference between the terminals is given by,

$$\begin{gathered} V=\epsilon -ir \\ V=12V-(50A)(0.040\Omega )=10V \end{gathered}$$

Hence, the potential difference between the terminals is 10V

(e) determining the rate energy dissipation

The current is 50A, then the rate P_rof the energy dissipation is as follows :

The rate energy dissipation is given by,

$P_r=i^{2}r=\left(50A{)}^2\right(0.040\Omega )=1.0\times {10}^{2}W.$

Hence, the rate energy dissipation is P_r=$1.0\times {10}^{2}W$

Therefore, use the formula for potential difference for a non-ideal battery to calculate the potential difference across the terminal of the battery and using power relation, find the energy dissipation inside the battery and the rate of energy conversion to the chemical form.