Battery

To run many of our household appliances we take the energy from a wall socket. We may also supply the energy from a battery, which continuously converts chemical energy into electrical energy and supplies it to the circuit to compensate for the energy lost to dissipation.

A battery is a device that holds a fixed potential difference V between its two terminals, A and B. The conventional assignment of + at terminal A and - at terminal B is indicated in the picture on the right. Chemical reactions in the electrolyte inside the battery drive positive ions to move towards the electrode A, and negative charges towards the electrode B. This results in an accumulation of positive charges on A, and negative charges on B. The accumulation process stops when the force on a charge due to the potential difference in the electrolyte resulting from this accumulation exactly equals the force due to the chemical reaction driving it to the electrodes.

When a wire is connected between A and B, this potential difference creates a field in the wire, moving electrons from B to A through the wire. For each electron that moves from B to A through the wire and neutralizes one unit of positive charge on A, the chemical reaction supplies a positive ion to A, and the process keeps going on and on.

The open-circuit voltage, i.e. the potential difference between A and B when no current is drawn by connecting the wire, is called the electromotive force, or emf. We denote it by e. When a current flows through the battery, the terminal voltage = VA - VB is usually less than e. This is because of the internal resistance, rb, of the battery. From the diagram, it is clear that the resistance of the wire (and whatever devices connected to the wire), R, and rb are in series (the same current I flows through them). Then

$\epsilon$ = I $\cdot$ ( R + rb )

which means that the terminal voltage

V = I $\cdot$ R = $\epsilon$- I$\cdot$ rb

Example:

Typical flashlight batteries have emf. = 1.50 V, but a relatively high internal resistance, say rb = 3 $\Omega$. So when a current of 50 mA is drawn from it, its terminal voltage will be 1.50 V - (50$\cdot$ 10-3 A)$\cdot$ ( 3 $\Omega$ ) = 1.50 - .15 V = 1.35 V. However, a car battery, made from lead and acid, has a rather low internal resistance, say rb = 0.01 $\Omega$. Its emf of a single lead/acid cell is usually 2.00 V, so even when a relatively high current of 5 A is drawn, it gives a terminal voltage of 2.00 - (5$\cdot$ 0.01) V = 1.95 V.

However, for most purposes in this class, the distinction between the terminal voltage, V, and the electromotive force, $\epsilon$, is not important, and we will usually use V instead of $\epsilon$.

Points to remember:

  1. A battery has two important characteristics : emf., $\epsilon$, and internal resistance rb.
  2. The emf. is determined by its chemical constituents : the material of the two electrodes, and the material of the electrolyte.
  3. The internal resistance is determined by features of its construction, like the distance between the plates, and the common area of the plates facing each other.
  4. The maximum terminal voltage one can get from a battery is its emf., $\epsilon$.
  5. The maximum current that a battery can deliver is its short circuit current (i.e., current when the external resistance is made 0), which is $\epsilon$/ rb .
  6. Caution: the symbol that one uses for the electromotive force, $\epsilon$, is the same as that for the permittivity. Please make sure that you do not confuse these two quantities.

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