Electrochemistry: cell vs battery

Tutoring chemistry, the distinction between cell and battery is noted.

In electrochemistry, a cell is a single unit of electrical energy production. A cell comprises an anode and cathode, plus the ingredients and the environment needed for the chemical reaction that outputs electrical energy.

A battery comprises more than one cell connected so that they work together to deliver energy to a circuit.

People have come to refer to single cells as batteries. I’d say that the button-style power sources found in calculators, watches, etc are cells. If a calculator contains two of them, those two cells constitute a battery.

The typical car battery really is one, since it contains six cells connected.


Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Giancoli, Douglas C. Physics, fifth ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Circuit analysis: The maximum power transfer theorem

From time to time, a tutor might get asked questions about electric circuits.  In the context of tutoring or just for general interest, the maximum power transfer theorem  is nice.

I last studied electronics about twelve years ago.  Ideas from it return to mind now and then.  Researching for my Nov 10 article on auto batteries, I read a remark that whatever the battery’s resistance was, the starter should match it to receive maximum possible power. I recognized the idea as a case of the maximum power transfer theorem:

Whatever the resistance of the surrounding circuit, the load resistor should match it in order to receive maximum power.

Here’s a proof using calculus:

Let’s imagine a series circuit with peripheral resistance R and load resistance aR, where a \geq 0. Since the two resistances are in series, R_{total}=R+aR.

Now, since V=IR, we have  I=\frac{V}{R}. In particular,


Furthermore, the power dissipated by a resistor is given by P=I^2R. Therefore, the power in the load resistor of our circuit is

    \[P=(\frac{V}{R+aR})^2 \times aR=\frac{aV^2R}{(R+aR)^2}\]

To find the value of a that gives the maximum value of P, we take the derivative  \frac{dP}{da}, set it equal to zero, and solve for a.

To take the derivative  \frac{dP}{da}, we use the quotient rule:

    \[\frac{dP}{da}=\frac{(R + aR)^2V^2R - aV^2R \times 2(R + aR)R}{(R + aR)^4}\]

We set \frac{dP}{da} to zero and solve for a:

    \[0=\frac{(R + aR)^2V^2R - 2aV^2R^2(R + aR)}{(R + aR)^4}\]

Multiply both sides by (R + aR)^4

    \[0=(R + aR)^2V^2R - 2aV^2R^2(R + aR)\]

Factor R + aR:

    \[0=(R +aR)[(R + aR)V^2R - 2aV^2R^2]\]

Divide out R + aR from both sides:

    \[0=(R + aR)V^2R - 2aV^2R^2\]

Factor out V^2R:

    \[0=V^2R[(R + aR) - 2aR]\]

Divide out V^2R:

    \[0=R + aR - 2aR\]

Factor out R:

    \[0=R(1 + a -2a)\]

Divide out R:

    \[0=1 + a -2a\]


    \[0=1 -a\]



Let’s recall that, in our circuit, the peripheral resistance is R, while the load resistance is aR. We now find that for maximum power, a=1. It follows that the load resistance should be 1R=R, the same as the peripheral resistance, for maximum power.

The maximum power transfer theorem, while many never encounter it, is a fundamental part of everyday life for many others. Anticipating what we may need to know in the future is often a challenge….

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.