Question

A particular 12 V car battery can send a total charge of 84 A-h (ampere-hours) through a circuit, from one terminal to the other.
(A) How many coulombs of charge does this represent?
(B) If this entire charge undergoes a change in electric potential of 12 V, how much energy is involved?

Answer 1

Even though the unit of charge in SI unit is coulomb, the charge produced by a battery is commercially measured in terms of the charge delivered by the battery in an hour and this quantity is known as ampere-hour (or Ah).

Complete step by step answer:
The charge is a characteristic property of a particle that causes it to exert force on another charge in its vicinity. The charge is a scalar quantity with two types namely, positive and negative charges. The SI unit of charge is coulomb (C).
Current is defined as the charge flowing through an area of cross-section per unit time. The current is measured by the SI unit ampere. If one coulomb of charge passes through an area of cross-section in one second, the current is said to be one ampere.
$1A = \dfrac{{1C}}{{1\sec }}$
(A) How many coulombs of charge does this represent?
Now, the charge can also be represented by the product of current in ampere and time in seconds. This gives us another commercial unit for measuring charge produced by a battery, known as A-h or ampere-hour.
If one ampere of current is supplied from the battery in one hour, the amount of charge discharged is said to be equal to one ampere-hour or 1 A-h.
To convert one A-h to coulomb, the A-h has to be converted to a product of ampere and second, which is equal to one coulomb. So,
$\Rightarrow 1Ah = 1A \times 3600\sec = 3600A - \sec = 3600C$
Therefore, if the battery produces the charge 84 Ah, the charge produced in coulombs is given by –
$\Rightarrow 84Ah = 84 \times 3600C = 302400C$
Thus, the battery produces a charge of 302400 C or $0 \cdot 30 \times {10^6}C$

(B) If this entire charge undergoes a change in electric potential of 12 V, how much energy is involved?
The power dissipated in an electric circuit is given by the product of voltage and current in the circuit.
$\Rightarrow P = VI$
Since, power is the rate of work done per unit time, the energy dissipated is –
$\Rightarrow E = Pt = VIt$
Given, voltage of the battery, $V = 12V$ and the product of the current and time is equal to charge.
$\Rightarrow E = 12 \times It$
$\Rightarrow E = 12 \times C$
Given that the charge is equal to 302400 C, the energy involved is given by –
$\Rightarrow E = 12 \times 302400 = 3628800J = 3.63MJ$

The energy dissipated is equal to 3.63 MJ.

Note: Even though it may seem by logic that the unit of charge, coulomb is a basic unit and ampere is derived from the unit of charge, in actual, ampere is one of the 7 basic units prescribed by the International System of Units and the unit coulomb is derived from the units of current and time, ampere and second respectively.