3 Physics and math
Right, now we've established the notation, here we go. For all electric equipment to operate, a voltage is needed. For example, a wall outlet supplies 115 Volts (US) or 230 Volts (Europe), and a car battery 12 volts. Once a device is connected (and switched on) current will flow, and power is generated. The amount of power is dependent on the voltage of the source and the resistance of the load. Before we start to describe everything in detail, we first we have to make some conventions:
Voltage, current and power
The relationship between voltage, current and power is best illustrated by this simple equation:
|P = V · I||Power equals Voltage times Current. Math is easy ain't it?|
Example 1: What's the current through an electric heater that's 115 Volts, 1750 Watts?
Filling in the equation P = V · I
gives 1750 W = 115 V · I
so I = 1750 W / 115 V = 15.2 A
Example 2: What's the maximum power generated by a car generator, rated 14.4 V, 70 A?
P = V · I = 14.4 V · 70 A = 1008 W
Resistance and impedance
Resistance is another basic aspect of electricity. It determines how much current will flow though a certain load at a certain voltage level. To be more precise: 1 Volt is generated when 1 Ampère flows through a 1 Ohm load. Calculation with resistances is much the same as with loudspeaker impedances [REF(@p:0303)]. In an equation it looks like this:
|V = I · R||Voltage equals current times resistance|
Example: What's the current through a 10 Ω resistor connected to a 12 V battery?
V = I · R
gives 12 V = I · 10 Ω
so I = 12 V / 10 Ω = 1.2 A
Now for some shortcuts. If we combine the two above equations we get:
|P = I2 · R||( P = V · I, substitute V by I · R )|
|P = V2 / R||( P = V · I, substitute I by V / R)|
Example (tough one): What's the current through a 10 kΩ resistor connected to a 250 mV (line level) source, and how much power is generated (just to show how low power line level signals really are)?
V = I · R
I = V / R
I = .25 V / 10,000 Ω
I = .000025 A = 25 μA
P = V · I
P = .25 V · .000025 A
P = .00000625 W = 6.25 μW
Calculating series resistances isn't very hard. You can just add up their values to find the substitute value Rs:
|Rs = R1 + R2 + R3 + ... + Rn||(n resistors in series)|
It so happens to be when you connect two identical resistors in parallel, their combined resistance is half that of one.
With parallel connections of different resistors, it gets more complicated: say you have two resistors R1 and R2. Their substitute value Rs will be:
|Rs = ( R1 · R2 ) / ( R1 + R2 )||(2 resistors in parallel)|
The following is a more general, but also more complicated, equation, in which you can put as many resistors as you like in parallel:
|Rs = 1 / ( 1 / R1 + 1 / R2 + 1 / R3 + ... + 1 / Rn ))||(n resistors in parallel)|
A useful help, when working with these equations, is to put them in triangles, like this:
|V I||I R||I2 R||P R|
Knowing two variables out of three, you can find the missing one by multiplying or dividing them. If the two known numbers are horizontal, multiply them. If they're vertical, divide the top number by the bottom one.PREV NEXT
© Joris van den Heuvel 2001-2009