# 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.

Source:

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.

# 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≥0. Since the two resistances are in series, Rtotal=R+aR.

Now, since V=IR, we have I=V/R. In particular,

I=V/(R+aR)

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=(V/(R+aR))^2(aR)=aRV^2/(R+aR)^2

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

To take the derivative dP/da, we use the quotient rule:

dP/da=((R + aR)^2(RV^2) – aRV^2(2(R + aR)R))/(R + aR)^4

Set dP/da to zero and solve for a:

0=((R + aR)^2(RV^2) – aRV^2(2(R + aR)R))/(R + aR)^4

Multiply both sides by (R + aR)^4

0=(R + aR)^2(RV^2) – 2aV^2R^2(R + aR)

Factor R + aR:

0=(R +aR)[(R + aR)RV^2 – 2aV^2R^2]

Divide out R + aR from both sides:

0=(R + aR)RV^2 – 2aV^2R^2

Factor out RV^2:

0=RV^2[(R + aR) – 2aR]

Divide out RV^2:

0=R + aR – 2aR

Factor out R:

0=R(1 + a -2a)

Divide out R:

0=1 + a -2a

Simplify:

0=1 -a

Finally,

a=1

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.