Finding the formula for electric energy and power

In an electrical circuit, the potential difference (voltage) – denoted by \(\text{U}, \text{V}\) or \(\Delta \text{V}\) is defined as the work done per unit charge:

\[\Delta \text{V} = Wq\]

The intensity of the electrical current is defined by the charge traveling through a conductor section in a unit time:

\[I = \frac{q}{t}\]

Electrical power is the amount of electrical energy generated per unit time. That electrical energy is the work done on the electrons (by the generator). So, we have the following formulas:

\[P = Wt = \frac{Uq}{q/I} = UI\]

From Ohm’s law, \(P = UI = RI^2 = \frac{U^2}{R}\)

Electrical energy (W) is \(W = P \Delta t = UI \Delta t = RI^2 \Delta t = \frac{U^2}{R} \Delta t\) and (obviously) it depends on the time the circuit has produced energy for.

Conversion of electrical energy by the consumers

Consumers in the circuit convert the electrical energy we have calculated above into other forms of energy. Light bulbs into light energy, resistors usually into heat and motors into mechanical energy.

This process doesn’t usually have a 100% efficiency, as some of the energy is lost (lost doesn’t actually mean the energy disappears, it means it is converted into an undesired energy form).

We define the efficiency of a consumer as the ratio between the useful energy produced by it and the electrical energy the circuit “loses” in the consumer:

\[\eta = \frac{E}{W}\]

Simple DC circuit

Consider a source of e.m.f. E, internal resistance r and a resistor of resistance R connected to it. By Ohm’s law, the current in the circuit is \(I = \frac{E}{R+r}\)

The electrical power in the resistor, using the formula above is

\[P= RI^2= R(\frac{E}{R+r})^2\]

Similarly, the power lost in the source is \(p=r(\frac{E}{R+r})^2\)

The total power is \(P_{tot} = P + p = \frac{E^2}{R+r}\)

We define the efficiency of the circuit as \(\eta = \frac{P}{P_{tot}}\)

We can show \(\eta = \frac{R}{R+r}\)

For the special case R=0, the efficiency is 0, and for the special case R→\(\infty\), the efficiency is 1. For any other value of R, the efficiency will be somewhere between 0 and 1.

Maximum Power Delivered to a Resistor by a Given Source

Suppose we have a source with EMF \( E \) and internal resistance \( r \), and we can choose the external resistance \( R \) to maximize the power delivered to it.

The power delivered to the resistor is given by: \[ P = \frac{R E^2}{(R + r)^2} \] Since \( E \) is constant, we aim to maximize the expression: \[ \frac{R}{(R + r)^2} \]

Let’s use completing the square to find the maximum of this expression. Let’s set: \[ f(R) = \frac{R}{(R + r)^2} \] Let’s define \( x = R \). Then, \[ f(x) = \frac{x}{(x + r)^2} \] Now write the denominator as a perfect square: \[ f(x) = \frac{x}{x^2 + 2xr + r^2} \] To maximize this, we consider the function: \[ f(x) = \frac{x}{x^2 + 2xr + r^2} \] and use calculus to find the maximum.

Take the derivative: \[ f'(x) = \frac{(x^2 + 2xr + r^2)(1) - x(2x + 2r)}{(x^2 + 2xr + r^2)^2} \] Simplify the numerator: \[ f'(x) = \frac{x^2 + 2xr + r^2 - 2x^2 - 2xr}{(x^2 + 2xr + r^2)^2} = \frac{-x^2 + r^2}{(x^2 + 2xr + r^2)^2} \] Set \( f'(x) = 0 \), so: \[ -x^2 + r^2 = 0 \Rightarrow x = r \] Therefore, the power is maximized when \( R = r \).

Substituting back: \[ P_{\text{max}} = \frac{r \cdot E^2}{(r + r)^2} = \frac{r E^2}{4r^2} = \frac{E^2}{4r} \]

At this point, the efficiency of the circuit is: \[ \eta = \frac{P_{\text{external}}}{P_{\text{total}}} = \frac{1}{2} \] So even though this is not the most efficient setup, it is the one where the external resistor receives the maximum power.

Example 1: Finding the efficiency Example 2: EMF and the internal resistance of a souce Example 3: Finding the internal resistance

Written by Alex Jicu