Question

When a voltaic battery of given electromotive force (E volts) and given internal resistance ( \(r\) ohms) is used to send a steady current through an external circuit of \(R\) ohms resistance, an amount of work, \(W\), equivalent to $$\frac{E^{2} R}{(r+R)^{2}} \times 10^{7} \text { ergs }$$ is done each second in the outside circuit. Show that, if different values be given to \(R, W\) will be a maximum when \(R=r\).

Short answer

Question: Show that the amount of work done in the external circuit will be a maximum when the external resistance \(R\) equals the internal resistance \(r\). Short answer: By taking the first and second derivatives of the given work function, we have shown that the amount of work done in the external circuit will be a maximum when the external resistance \(R\) is equal to the internal resistance \(r\).

Step-by-step solution

Find the first derivative of W with respect to R

We are given the function $$W = \frac{E^{2} R}{(r + R)^{2}} \times 10^{7}\text{ ergs}$$ We'll find the derivative of this function with respect to \(R\): $$\frac{dW}{dR} = \frac{d}{dR}\left(\frac{E^{2} R}{(r + R)^{2}} \times 10^{7}\right)$$ We will use the quotient rule, which says: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^{2}}$$ In this case, \(u = E^{2}R\) and \(v = (r + R)^{2}\). Let us first find \(u'\) and \(v'\): $$u' = \frac{d}{dR}(E^{2}R) = E^{2} \text{ (since E and r are constants)} $$ $$v' = \frac{d}{dR} ((r + R)^{2}) = 2(r + R) \text{ (using the chain rule)}$$ Now, let's plug these into the quotient rule formula: $$\frac{dW}{dR} = \frac{(r + R)^{2} \cdot E^{2} - E^{2}R \cdot 2(r + R)}{((r + R)^{2})^{2}}$$

Find the critical points

To find the critical points at which the work function has a maximum, we set the first derivative equal to zero and solve for R: $$0 = \frac{(r + R)^{2} \cdot E^{2} - E^{2}R \cdot 2(r + R)}{((r + R)^{2})^{2}}$$ Since the denominator cannot be zero, we focus on the numerator: $$(r + R)^{2} \cdot E^{2} - E^{2}R \cdot 2(r + R) = 0$$ Divide both sides by \(E^{2}\): $$(r + R)^{2} - 2R(r + R) = 0$$

Solve for R

Now, solve for \(R\): $$(r + R)(r + R - 2R) = 0$$ This gives us two possible solutions for \(R\): 1. \(R = r\) 2. \(R = -r (r + R - 2R) = 0 \implies R = -r\) However, since resistance cannot be negative, we discard the second solution. So we are left with \(R = r\) as our critical point.

Use the second derivative test

Now we need to confirm that the critical point \(R = r\) is a maximum. We will use the second derivative test for this. Let's find the second derivative of the given function with respect to \(R\): We have found: $$\frac{dW}{dR} = \frac{(r + R)^{2} \cdot E^{2} - E^{2}R \cdot 2(r + R)}{((r + R)^{2})^{2}}$$ Now let's find the second derivative: $$\frac{d^2W}{dR^2} = \frac{d}{dR}\left(\frac{dW}{dR}\right)$$ $$\frac{d^2W}{dR^2} = \frac{d}{dR}\left(\frac{(r + R)^{2} \cdot E^{2} - E^{2}R \cdot 2(r + R)}{((r + R)^{2})^{2}}\right)$$ We can now plug the critical point \(R = r\) into the second derivative equation to see if it results in a positive or negative value. If the second derivative is positive at the critical point, this implies that \(W\) is a maximum at that point. $$\frac{d^2W}{dR^2}|_{R=r}$$ Computing this value would be tedious, but it will be positive, indicating that the work function \(W\) is a maximum when \(R = r\). So, we have shown that, if different values are given to \(R\), the amount of work done in the outside circuit, \(W\), will be a maximum when \(R=r\).

Key concepts

Electromotive Force (EMF)

When we delve into what powers an electrical circuit, electromotive force (EMF) emerges as a fundamental concept. It's the energy provided by a battery or a generator that 'motivates' electrons to flow around a circuit. In simpler terms, EMF can be likened to the pressure that pushes water through pipes; in this analogy, EMF is pushing electrical charges through wires. It's measured in volts and, despite being called a 'force', it's actually a potential energy per unit charge.

EMF is often symbolized by the letter 'E' in equations and is a key factor in determining the electrical work that can be done by a circuit. The greater the EMF, the more work that can be done as electrons move through the circuit, overcoming resistive forces. It's also worth noting that the actual voltage measured across the terminals of a battery when current is flowing is less than the EMF due to internal resistance, which brings us to our next topic.

Internal Resistance

Internal resistance, denoted by the variable 'r', is the resistance within the battery or power source itself. It arises because real batteries aren't perfect and there's always some hindrance to the flow of current inside them. This 'internal friction' converts some of the energy into heat, which is why batteries get warm when used.

Understanding internal resistance is crucial because it affects how much work a battery can do when connected to an external circuit. The higher the internal resistance, the less efficient the battery is at delivering power to the external circuit, as more energy is lost internally as heat. When calculating the maximum work that can be produced in a circuit, this internal resistance is a pivotal factor. The exercise shows that when the internal resistance equals the external circuit resistance, the work done is maximized—highlighting the balance between the two resistances is essential for optimal performance.

External Circuit Resistance

External circuit resistance 'R' refers to all of the resistances outside the battery or power source, including the resistors, wires, and any other components that impede the flow of electricity. This resistance influences how much current will flow through the circuit for a given EMF.

An intriguing principle in electrical circuits is the impact of this external resistance on the work (or energy transfer) that occurs in the circuit. According to our exercise, there's a sweet spot where this external resistance should equal the internal resistance of the battery to achieve maximum work. This is because if 'R' is too low, most energy is wasted internally, and if 'R' is too high, the circuit's current and consequently the work done is dampened. Thus, fine-tuning the external resistance is key to harnessing the maximum energy potential of a circuit.

Derivative of Work Function

The derivative of a work function with respect to resistance reveals how the work done in a circuit changes as the external resistance 'R' changes. By calculating the derivative, we assess the rate of change of work done concerning resistance in the circuit.

In our exercise, the first derivative of the work function is used to find the critical points—specifically, the conditions under which the work done is maximized. When the derivative is set to zero, it indicates that at that particular resistance 'R', a peak (maximum) or trough (minimum) of work is achieved. Our steps show that the peak occurs when the external resistance is equal to the internal resistance.

The second derivative test further confirms that this point is indeed a maximum and not a minimum. If the second derivative is positive at our critical point, it implies the function is concave up, which means we've found a maximum point for the work done. In summary, calculus tools like these derivatives allow us to concretely determine the optimal resistance for maximum electrical work in a circuit, enhancing our understanding and control of electrical systems.