Lesson 05 – Ohms Law part 2
There is a video for this lesson.
Welcome to to Electronics For Fun
Lesson 5 – Resistors in Parallel
As I introduce new terms, I have included a link to Wikipedia. Read ahead a little, and if you still need help, you can click on any orange word below for more information than you probably want. 🙂
What is a Parallel Circuit
In the previous lesson, I talked about a series circuit. The resistors were wired end to end. Here is a simple diagram to refresh your memory.
The two resistors are wired in series. If we were to remove one of the resistors, or if one were to burn out, the circuit would be open, and there would be zero current flowing thru either one. It would be like old time Christmas tree lights, when one burned out the whole string went dead.
In the parallel circuit shown below, both resistors are wired across the battery terminals. The total current flow thru the battery is equal to the current flowing in the resistor shown in the middle, plus the current flowing thru the resistor shown on the right.
If we remove one resistor, like in the circuit shown below, obviously the current stops flowing in that resistor. But unlike the series circuit, current will continue to flow, as before, thru the remaining resistor.
In the diagram below, I have added some meters again. This time we have 3 ammeters to measure current, and only 1 voltmeter. Let’s start with the voltmeter. Since the voltmeter is wired directly across the battery, as are the resistors, it will read battery voltage or 10 volts. Remember, for these lessons, you can ignore the minor resistance of the ammeter and consider it just a piece of wire. In just a moment we will calculate the current thru each resistor and the total current.
But first, let’s look at the circuit closer. So starting at the + terminal of the battery, the current goes thru the first ammeter, and down thru the first 1 kohm (1000 ohms) resistor, thru that ammeter, and back to the negative terminal of the battery.
But it also, at the same time, goes over and down thru the second resistor, thru its ammeter, and back to the – terminal of the battery. We have 2 circuits at the same time, in parallel.
We calculate the current for each resistor, one at a time. In this case they are both the same value, but they can be different as well.
The current (I) is voltage (V) divided by resistance (R). Since they are the same we only need to find the current flow thru one. So current = V/ R which is 10/1000 or one hundredth of an amp, or .01 amps. We have that much current flowing thru each resistor, so the total current flowing out of, and back into the battery is twice that or .02 amps. So the voltmeter would read 10 volts, the main ammeter would read .02 amps and each of the ammeters hooked to the resistors would read .01 amps.
I want to introduce two new terms if you are not familiar with them. In AC household current we usually talk about whole amps like 15 amps for a heater, or maybe 200 amp service coming into the house. In electronics, with smaller components, we usually are dealing with a fraction of an amp.
The first term is the milliamp. One milliamp is one thousandth of an ampere or .001 amps. So the .02 amps from above might also be written as 20 milliamps, or just 20 ma. The second term we sometimes see in electronics is the microamp. A microamp is a millionth of an ampere or .000001 amps, or just 1μa. Sometimes it’s just written as 1ua, without the Greek letter.
OK, so up until now, our parallel resistors have been the same value, but of course that is not always the case. The equation gets trickier here. In the diagram above, we have 2 different resistors. Let’s say that the resistor marked as R1 is 1kΩ (1000 ohms) and the other one marked R2 is 2kΩ (2000 ohms). In series resistance, we only needed to add them up to get the total resistance. So that equation was RT = R1 + R2 + R3 + etc. RT meaning Resistance Total or simply the Total Resistance.
For resistors in Parallel we get a new equation:
1/RT = 1/R1 + 1/R2 + 1/R3 + etc
This may seem a little intimidating at first. If you want to cheat, scroll down past the math. But if you want to learn the math involved, stick with me here. When we break it down, it isn’t too bad. Plus when there are only two resistors we have a shortcut we can use. Don’t let the math scare you away, you could also just buy an inexpensive meter and just measure across the resistors to get the total resistance. BUT, if you really want to know electronics, you should learn the math.
Again, the normal equation looks like this:
1/RT = 1/R1 + 1/R2 + 1/R3 + etc
Since there are only two resistors, we could use the shortcut that I will show you in a minute, but we need to learn this in case there were more than two. Since we are using the “long” method, even though only have two resistors, our equation would be”
1/RT = 1/R1 + 1/R2 Which, if we plug the known values in becomes:
1/RT = 1/1000 + 1/2000 (The 1000 and 2000 are the value of R1 and R2 in ohms)
Next we divide 1 by 1000 and get .001 and 1 divided by 2000 is .0005 so substituting we get:
1/RT = .001 + .0005 and then after adding we get 1/RT = .0015 and finally, we divide 1 by the .0015 which gives 666.67 ohms.
Or, since it is only two resistors, we can cheat a little and use this equation: RT = (R1 times R2) divided by (R1 plus R2) which becomes
RT = (1000 times 2000) divided by (1000 plus 2000) which becomes:
RT = (2,000,000) divided by (3000) which becomes:
RT = 666.67 ohms
You can cheat even more and use an online resistance calculator like the one at www.allaboutcircuits.com. You can Click Here to try it out.
No matter which way you choose to solve the total resistance value, here is a tip for you. In parallel resistors, the total resistance will always be less than the lowest value, and in series resistors, the total will always be more than the highest value.
Test your knowledge if you feel like it with a little test. No cheating now!
Q1: Two equal value resistors are in parallel across a 10 volt battery. The current flowing thru them is ___________.
Q2: Two unequal resistors are wired in parallel across a 10 volt battery. If one of them burns out, the current in the other one
Q3: Two unequal resistors are wired in parallel across a 10 volt battery. The voltage across the larger resistor is __________________________.