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Greetings:

Electricity can be confusing because we use names and terms that are
not familiar

to everyone. To answer your questions I’ll use the liquid cooling
system fin an

automobile and compare it to an electrical circuit. First I’ll explain
an electrical

circuit in words and then I’ll present some basic ways to calculate
voltage, current

and resistance using Ohm’s Law.

In an automobile cooling system the engine drives a water (or coolant)
pump

which produces water pressure at the output port of the pump.

In an electrical circuit we use a chemical reaction within a battery
as an electron

pump. There is an electrical pressure of extra electrons around the
atoms at the

negative (-) terminal of the battery and an equal number of atoms
with missing

electrons at the positive (+) terminal of the battery. **We call this
electrical pressure
voltage (volts for short)**.

In the automobile we connect a hose from the output of the water pump to the

engine block. We use a large diameter hose so that it has a low resistance to the

flow of water and the water does not lose pressure before entering the engine

block. A small diameter pipe will resist the water flow more than a larger pipe and

so the smaller pipe will have a greater pressure drop across it relative to a larger

diameter pipe. We measure the flow of water in the cooling system in gallons per

minute (liters/minute).

In an electrical circuit we use a metal wire with a sufficient diameter so that it does

not resist the flow of electrons into the circuit.

in electrons per second and call it amperes (amps for short)

The engine block has small holes and tubes and resists the flow of water through it

causing a pressure drop between the input and output of the engine block.

Electrical devices such as lamps, heaters, motors etc. resist the flow of electrons

through them.

bulb in ohms

input to the out put of the bulb. For a fixed value of current flow, the greater the

resistance is in ohms, the greater the voltage drop will be across the resistance.

In the car the water heated by the engine block then passes through a low

resistance hose and into the radiator for cooling. The radiator has a different

resistance to the water flow than the engine block has. The radiator also has a

pressure drop from the input to the output. We use a low resistance hose to

connect the radiator back to the input port of the water pump. We say that the

engine block, the hoses and the radiator are connected in series with the pump

because there is only one path through which the water can flow. Some of the

pump pressure drops through the block and the rest drops across the radiator. We

will assume that the hoses have a very small or zero pressure drop.

The electrical circuit can have a low resistance wire connected from the first bulb

to a second lamp. The second bulb may or may not have the same resistance as

the first lamp. Depending on the amount of the resistance in ohms, there will be a

voltage (pressure) drop across each bulb. The output of the second lamp is

connected by low resistance wire to the positive (input) terminal of the battery. We

call this a

flow from the output terminal of the battery back to the input terminal of the

battery.

In a series electrical circuit (or water circuit) the total resistance of the circuit is the

sum of all of the resistances hooked in series. Because wires usually, but not

always, have very low resistances , we often leave their very small restiance values

out of our calculations.

When we connect a number of bulbs in a series circuit to a battery, the battery

terminals see the sum of the resistances in the circuit. This value of total

resistances determines the current flowing from the battery through the circuit. We

can calculate this current by using one of the versions of

(1) Amps = Volts divided by resistance; V = A/R

Now that we have calculated the current flowing through the circuit we can

calculate the voltage drop across each device in the series circuit using a second

version of Ohm’s Law:

(2) Volts = Amps times Resistance; V = A * R

For example let us have two 6 volt bulbs with different current ratings.

Bulb A is rated at 6 volts and 0.05 amps

Bulb B is rated at 6 volts and 0.1 amps

Using a third version of Ohm’s Law we can calculate the resistance of the bulbs

from their ratings:

(3) Resistance = Volts divided by Amps; R = V/A

The Bulb A resistance would be R = 6v/0.05 amp = 120 ohms

The Bulb B resistance would be R = 6v/0.1 amp = 60 ohms

Now if we hook these two bulbs in series to a 12 volt automobile battery, the

battery sees the total resistance of both bulbs:

R = 120 ohms + 60 ohms = 180 ohms

Using equation (1) we can now calculate the current from the battery flowing

throughout the circuit:

A = 12v/180 ohms = 0.067 amp

Using equation (2) we can now calculate the voltage drop across each bulb:

Bulb A: V = 0.067 amp * 60 ohms = 4.0 volts

Bulb B : V = 0.067 amp * 120 ohms = 8 volts

Notice that the sum of the two voltage drops equals the battery voltage. It has to

the pressure drop must equal the pressure applied.

This example demonstrates that the circuit current and the related voltage drops

are determined by the total resistance of the bulbs and not by the ratings of the

bulbs. In the example Bulb B is 2 volts over its rating and it will be brighter, hotter

and will not last as long as its rating. Bulb A is 2 volts under its rating and it will be

dimmer, cooler and will last longer that its rating. If we had used 2 Bulb As or 2

Bulb Bs in the circuit, Ohms law would show that each bulb would have dropped 6

volts and would be operating at its rated values. However, this would be a very

special case.

One other equation is the power equation. Electrical power is what we pay for and

will determine the life of the battery.

(4) Power in watts = Volts times amps; P = V*A

Using (4) we can calculate the power dissipated by each bulb and the power

generated by the battery:

Bulb A: P = 4v * 0.067 amps = 0.268 watts

Bulb B: P = 8v * 0.067 amp = 0.536 watts

Battery: P = 12v * 0.067 amps = 0.804 watts

Notice that the power delivered by the battery equals the power dissipated in the

bulbs. It has to be the same!

Once you get used to considering volts as pressure , amps as current flow and

ohms as resistance to current flow you will have a greater understanding about the

operation of electrical and electronic circuits.

I should point out that we use parallel circuits in our homes, cars, airplanes etc.

because in a series circuit if one device blows out the whole circuit fails. Parallel

circuits are also analyzed by Ohm’s Law by considering each parallel branch as a

series circuit and then summing all of the currents in each series circuit at the

battery terminals.

Ohm’s Law is a very powerful tool and I have used it along with a simple $20

Volt-Amp--Ohm meter to trouble shoot and repair hundreds of automobiles,

radios, TV sets and aircraft and spacecraft electrical systems.

Best regards, Your Mad Scientist

Adrian Popa

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