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Bredhurst Receiving and Transmitting Society

ILC title


3. Technical Basics

3b Simple circuit theory

3b.1 Understand the relationship between power, potential difference and current. Manipulate the equation P = V x I to find the unknown quantity given the other two. The prefixes milli and kilo may be used.

Power in measured in the unit WATT or W and in the FLC you were introduced to a magic triangle this is also needed in the ILC so that you can manipulate the equation P = V x I to find the unknown item given the other two items.

power magic triangle

Practice with this until you are certain that you fully understand how to find one unknown from two know items.

NOTE: calculations in the exam may use the prefixes milli and kilo.

A student from an early course suggested the following to help you remember the order of the letters :-

P = Prefers V = Vanilla and I = Ice-cream

hence

Prefers Vanilla Ice-cream

Example :- What power supply would be suitable to run a 50W transceiver from a 12V power supply?

from P = V x I

we get 50 = 12 x I

then I = 50 / 12

result 4.166 Amps

The answers given in the exam might be :-

A 12V at 500mA

B 12V at 1A

C 12V at 6A

D 24V at 4A

D is wrong as it is at the wrong voltage, A & B are two low so the answer is C

However be aware that the output power actually bares no direct relationship to the input power as sometimes to have an output of 50W RF you need 100W of DC input !!!

Recall that a current through a resistor results in a transfer of electrical energy to heat energy in the resistor.

Because resistance is the opposition to the flow of electrons the energy of the electrons being slowed down results in the generation of heat in the resistor. This is why resistors have a resistance value and a power wattage to indicate how much power they can dissipate.

Understand the relationship between potential difference, current and resistance. Manipulate the equation V = I x R to find the unknown quantity given the other two. The prefixes milli and kilo may be used.

In the FLC you were introduced to a magic triangle this is also needed in the ILC so that you can manipulate the equation V = I x R to find the unknown item given the other two items.

Voltage magic triangle

Practice with this until you are certain that you fully understand how to find the one unknown from two know items.

NOTE: calculations in the exam may use the prefixes milli and kilo.

Again student from an early course suggested the following to help you remember the order of the letters :-

V = Vanilla I = Ice-cream R = Ready

hence

Vanilla Ice-cream Ready

Understand circuits comprising series and parallel connections of resistor and cells.

Calculate currents and potential differences in such circuits.

The circuit below Fig1 the Resistors R1, R2 and R3 are the opposition to current flow around the circuit to find the current flowing in the circuit one must first work out the total resistance value.

Fig1

To work out the two equal value resistor in parallel Rtotal = (R1+R2)/2

So Rtotal= (1000+1000)/2 Rtotal=500 ohms

Now to work out the total resistance RT in the circuit when resistor are in series add together

RT = Rtotal + R3

So RT= 500 + 12000 = 12500.

To find the total current flow from the battery and at points C and D using Ohms law.

Itotal = V1/ RT so Itotal = 12/12500 = 960uA

To workout the voltage drop crossed R3 calling it V3 using Ohms law

V3 = R3*Itotal

V3 = 12000*960uA = 11.52V

so V3 = 11.52 Volts

To workout the voltage drop across R1 and R2 which is V2 there are a number of ways.

First way because we have worked out the voltage drop across R3

V2= V1-V3

so V2= 12 - 11.52 = 0.48V

V2 = 0.48V.

The second way because we have worked out the two resistor in parallel R1 and R2 which was called Rtotal and also the total Current flowing in Rtotal is Itotal using Ohms law V2 = Rtotal x Itotal

So V2 = 500 x 960uA = 0.48V

So to sum up the voltage drop across R1 and R2 is V2 = 0.48v, the voltage drop across V3 is 11.52v

The current through R3 is 960uA

But what about the current flowing through R1 to work this out using Ohms law because the voltage drop across R1 has be worked out V2 the current flow through R1 is V2/R1

So 0.48/1000 = 480uA

To work out current flowing through R2 is the same way has working the current flowing in R1 so V2/R2 = 0.48/1000 = 480uA

Calculate the combined resistance of two or three resistors in series. Calculate the combined resistance of two or three equal resistors in parallel.

SERIES RESISTORS

Resistance is the opposition to current flow. The bigger the resistance the smaller the current that can flow.

Before we look at the topic in electronic terms let's consider a large number of people all lined up in one long straight line.

The effort that you would need, as an electron!!, to push through would be quite a bit but not impossible.

Now what would happen if the crowd broke into two lines you would now have to push your way through the first and then the second line and you would find it more difficult and you would have used up more energy.

So if the crowd broke into three lines you would now have to push your way through all three and that would be more difficult than the other two previous examples and even more energy would have been used up.

The same is the case with resistors. resistor

When there is just one resistor in a circuit as shown below then the amount of current flowing for a know voltage through a known resistor can be calculated based on the formula V = I x R and would be shown on the amp meter or ammeter (a ammeter in series with the circuit to measure current).

which re-arranged give you the formula shown below.

I = V / R

When we have two resistors in series, as shown below,

the current has to flow through both and finds it more difficult.

In fact the value of the total resistance is simply to add up the values of the individual resistors.

"SERIES Resistor TOTAL" = R1 + R2

Even though the two resistors are separated by the meter they are still connected in series.

So to find the current flowing in a circuit, through the meter, with two resistors in series you first work out the value of the total resistance and from that calculate the current from

I = V / R where R = R1 + R2

So when we have three resistors all lined up one behind the other the task of the current passing is still greater and less flows.

As with the two resistors mentioned above the total resistance is simply to add up the values of the individual resistors.

"SERIES Resistor TOTAL" = R1 + R2 + R3

Current flow

Above the three resistors are "in SERIES" as the current has to flow though each as there is no other way. also it does not matter where in the circuit the meter is to measure the current.

So to find the current flowing in a circuit with three resistors in series you first work out the value of the total resistance and from that calculate the current from

I = V / R where R = R1 + R2 + R3

Voltage across each resistor

If the voltage is measured across each resistor then the sum of these individual values will be the same as that across the battery as the set of resistors could be replaced by a single resistor equal to the total value of the individual resistors (as they are in series)

PARALLEL RESISTORS

The other way that resistors can be coupled together is for each end of one resistor to be linked to the same end of the next resistor and the same with the other end.

We now have a different situation to the "SERIES" as the electrical path, as we call it, has two routes to chose from. If each of the resistors is of equal value then it is as difficult for the current to pass through each resistor and therefore the current divides equally between the resistors.

If they are not of similar value ....... well that is for a later training course so let's just think in terms of similar values

R1 = R2

Value of Two EQUAL resistors in parallel = Value of one resistor divided by 2

"Parallel Resistance Total" = R1 / 2

R1 = R2 = R3

Value of Three EQUAL resistors in parallel = Value of one resistor divided by 3

"Parallel Resistance Total R" = R1 / 3

This may look strange but if you think of it this way .... there are in the first example two paths that the current can take therefore with equal values of resistor the resistance is halved as there are two paths

and with three resistors of equal value the resistance is a third as there are three paths!!

Students are not understanding that when faced with a question of a current in a simple circuit that you must first determine the value of the total resistance whether that is resistors in series (as outlined above) or resistors in parallel as outlined below. It is only then that you can go forward to calculate the current flowing for a particular applied voltage but the use of Ohm's law. so let's look in a bit more details about this type of question.

Current in the circuit

Find the current in a circuit with resistors in parallel

You must first find the actual total amount of resistance and from that calculate the current flowing in the circuit. The meter measure the total amount in the circuit and MUST be placed to make this reading outside the resistor network else it would only read the current passing through the single resistor with which it is in series.

Practical example - Make a DUMMY LOAD.

Let's take a practical example. If you want to make a dummy load then how many carbon resistors will you need and of what value to make a 40W dummy load ?

There would be several ways of going about this :-

50 0.8W 1 ohms resistors in series. That would make a fairly bulky dummy load and probably only 1W resistors available so the unit would be over rated.

So look at it another way. 2W resistor of the carbon type are available and thus 20 resistors would give us the power capability we require and assuming we are using them in PARALLEL, what value of resistor is needed ?

From the equation above :-

Resulting value = Value of one resistor / Number of resistors

rearranging the equation we get :-

Value of one resistor = Resulting value x Number of resistors

Value of one resistor = 50 x 20 = 1000 ohms or 1k ohm

So to make a simple dummy load we need 20 1k 2w resistors linked in parallel.


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