Time for some simple equations. Not to worry, equations are just relationships or another way for saying how something is dependent on another. If you turn up the heating you will sweat. The more you turn up the heating the more you’ll sweat. The sweating is a function of heat. It is very similar with Power.

Charge

We talked about charge in an earlier topic. An atom consists of a nucleus and electron
around it. Usually, the number of positively charged protons in the nucleus is the same as the number of electrons.

  • Atom is positively charged when it loses electrons
  • Atom is negatively charged when it gains extra electrons
Equation: Current is the charge over time

Where I is the current, Q is the charge and t is time.

So, the current is equal to the charge over time. Remember, if 6 million million million electrons pass per second I = 1A.

Electrical Energy

  • Energy comes from the battery and is supplied to the charges (the electrons). This is where the electrons get the energy from to jump lose from the atom.
  • Electrons deliver this energy to the load (light bulb) and go back to the battery for more.
  • The bulb converts this energy to light and heat
  • Electrical Energy is measured in joules (J)
Equation: relationship between voltage, electrical energy and charge

Where W is the Electrical Energy, V is the voltage and Q is the charge.

What is 1V?

Let’s have that light bulb as an example. Electrons carry the electrical energy. So, they pass through the bulb and give that energy to the bulb so it can transfer it into other forms of energy.

If 1 Coulomb of electrons, that is 6 million million million electrons pass through the bulb and deliver 1 joules of electrical energy to be transferred into other forms of energy, the voltage across the bulb is 1V.

1V = 1 Joule/Coulomb

Power

Earlier we mentioned, that the electrical energy electrons carry, is converted into other types of energy such as heat or light.

  • Power is the rate this electrical energy gets converted at.
  • The unit of power is the watt (W)
  • 1W is when 1J of electrical energy is converted to other form of energy.

1W = 1J/s

Equation: Power equals the current times the voltage

Where:

  • P is the power
  • I is the current through the bulb
  • V is the voltage across it.

Example:

Taking our battery circuit for an example. In the circuit below we have a battery supplying 5V to the bulb and there is 1A in the circuit.

Therefore, the power is: P = 5×1 = 5W. Since 1W = 1J/s, this bulb converts 5J of electrical energy into light and heat.

Example circuit for calculation power

There are two more equations worth mentioning:

Equations for calculating Power
  • It is true only in the case of a resistive element (Resistor) that changes all electrical energy to heat.
  • We know from before that Ohm’s Law is R=V/I. By rearrange it we get V=IR.
  • To get the first equation, simply substitute V=IR to P=IV. So you get P=IxIR.
    IxI is I^2 so P=(I^2)xR
  • To get the second equation, rearrange Ohm’s law to get I and substitute to P=IV.

From the first equation, the Power is directly proportional to R only if the current through the resistor is constant.

From second equation, P is inversely proportional to R if the voltage across the resistor is constant.

Maximum Power Transfer

The question is: what value should the load resistance (RL) be so we have the maximum power transferred for a fixed Rs?

In the following circuit, we have 3 elements: a voltage source, the internal resistance of the voltage source and a load.

The short answer is: For maximum power transfer the load resistance has to be equal to the internal resistance of the battery.

Let’s prove this.

Example Circuit for Maximum Power Transfer

Max Power Transfer Calculation

To find the value of RL let’s start with 2 equations that we are familiar with:

Calculating Current in the Maximum Power Transfer example circuit
Calculating the Power dissipating in the load resistor

Simple. Equation 1 is from Ohm’s Law V=IR. Rearranging, we get: I=V/R. I is the current flowing through the circuit, also though RL. Vs (source voltage) in Eq 1 is the same as V in Ohm’s equation, so Vs=V. We get R from adding up Rs and RL since they are in series.

Substituting the first equation into the second we get:

The Characteristic Curve

What we want is to get the maximum value of P by varying RL.

From this we can draw the characteristic curve of P vs RL. Don’t worry if you don’t know how to do this. Just accept that this is done by a computer by comparing how P and RL vary with different values.

The characteristic curve of P vs RL

The curve looks like an inverted U and tells us:

  • By increasing RL from 0 initially increases the power
  • There is a certain value of RL where P reaches it maximum
  • As RL increases further P drops.

Here comes a bit of differentiation:

We can alway draw a tangent to any point on a curve which is essentially a straight line. Now, analysing this curve, at the point where P is maximum, the tangent has a slope of 0. Slope 0 means it is completely horizontal. In the world of mathematics, we express is as:

Expression that shows where P is maximum the tangent has a slope of 0

Differentiations has some straight forward rules. Here are a few:

  • (1)      \dfrac {d}{dx}x=1

  • (2)      \dfrac {d}{dx}x^{n}=nx^{n-1}

  • (3)      \dfrac {d}{dx}\left( upm v\right) =\dfrac {du}{dx}pm \dfrac {dv}{dx}

But when it gets a little more complex there are more complicated rules such as the Quotient Rule:

if y=\dfrac {u}{v} then \dfrac {dy}{dx}=\dfrac {v\dfrac {du}{dx}-u\dfrac {dv}{dx}}{v^{2}}

Here is the differentiation bit again:

\dfrac {d}{dR_{L}}\left[ V^{2}_{s}\dfrac {R_{L}}{\left( R_{L}+R_{s}\right) ^{2}}\right] =0

 

V^{2}_{s}\left[ \dfrac {\left( R_{L}+R_{s}\right) ^{2}-R_{L}\left[ 2\left( R_{L}+R_{s}\right) \left( 0+1\right) \right] }{\left( R_{2}+R_{s}\right) ^{4}}\right]  =0

Steps to solve the above differentiation:

1 – We are differentiating in respect of RL. Vs^2 is a constant (no RL in it) so we write that down

2 – From the Quotient Rule let’s say u=RL and v=(RL+Rs)^2. First we are focusing on how to get the numerator in the brackets.

    • According to the Quotient Rule, in the numerator we write down v first which is multiplied by du/dx.
    • du/dx is general. In our case we are differentiating in respect of RL so dx = RL. Since u=RL as discussed above by substitution we get: dRL/dRL = – using the first rule above.
    • next, we subtract u times dv/dx. u=RL so we write that down. dv/dx means we differentiate (RL+RS)^2 in respect of RL.
      • According to a rule we treat this as X^2 first, differentiate it in respect of X, and we multiply the result with the derivative of what’s inside the brackets.
      • by definition dX^2/dX = 2X. So, the first line will be 2(RL+Rs) and we multiply this with the derivative of (RL + Rs) in respect of RL (3rd rule above)
      • dRL/dRL = 1. Since we are differentiating with respect of RL: dRs/dRL = 0 as Rs is considered as a constant in this case.

3 – The denominator is straight forward. Since according to the Quotient Rule, it is v^2 and v=(RL+Rs)^2 the denominator will be (RL=Rs)^4

Following some simple algebra, we get to the result that the maximum power dissipating happens if the source resistance is equal to the load resistance.

optional reading: Success in Electronics book by Tom Duncan
NEXT TOPIC: Kirchhoff’s Law