Show that a battery delivers the most power when the load resistance across its terminals is equal to its internal resistance. (This is not the way to treat a battery, but it's the basis for load matching in amplifiers; see Problem \(65 .\) )

Short Answer

Expert verified
It has been proven that the maximum power is delivered by a battery when the load resistance equals the internal resistance. This was obtained by deriving the power expression in respect of Load resistance and setting it to zero.

Step by step solution

01

Represent given parameters

Let's denote the internal resistance of the battery as \(r\) and the load resistance as \(R\). The battery produces an electromotive force, \(E\).
02

Set up the equations

The current in the circuit, according to Ohm's law, can be expressed as \(I=E/(r+R)\), and power (P) delivered to the load is given by \(P=I^2.R\). We can substitute the expression for I from the current equation into the power equation, and we get: \(P=E^2.R/(r + R)^2\).
03

Find the maximum power

To find the maximum power, we should derive the power expression with respect to the load resistance R and equate it to zero, as the maximum or minimum of a function occurs where its derivative is zero. Doing the necessary differentiation, we get: \(dP/dR=(E^2)/(R+r)^3 - 2*E^2*R/(R+r)^3\). Set this to 0 and solve for R. You will find that \(R=r\).
04

Check the validity of the result

To ensure that the value obtained is for maximum power and not minimum, we double differentiate the power equation and substitute R=r. If the second derivative is negative, it means we have a maximum point, which is what we should expect. The double differentiation of P in respect of R gives \(d^2P/dR^2=-2*E^2/(R+r)^3 + 6*E^2*R/(R+r)^4\). If we substitute R=r in the above equation we get it to be negative confirming that it is a maximum point.

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