- #1
I know that in series the current is the same all the way around the circuit, while in parallel it is a portion.
Rt = R1 + R2... etc ...
... so I = V/R, 10/ (Rn/Rt)
Zryn said:I = V/R is Ohms Law rearranged, but what does the 10/ (Rn/Rt) relate to?
Instead of jumping into the calculations, can you 'see' (describe) the method by which you would obtain the current through R4?
The 10 is voltage and Rn/Rt was supposed to be Rtotal but I wrote it wrong.
Zryn said:The voltage source in the diagram is labelled as v1 = 12V, not 10, and I wanted you to write out the equations that got to Rn/Rt so we could see where you were going wrong, whether it was with your method or your math .
There are several methods you could use to determine the current in R4, but here are two:
a) Combine all the resistors into one resistance, calculate the series current, and then use a current divider to determine the current in R4
b) Combine each set of parallel resistors into one resistance, calculate the voltage divider and then use Ohms Law to determine the current in R4
Which looks preferable to you? (Ideally you could do both to assist your knowledge and confirm your answers)
I have already written the equation above so would just need to add them all together for Rtotal?
Zryn said:Using Rt = 1/(1/R1 + 1/R2 + ... + 1/Rn) will allow you to find the total resistance of resistors in parallel, but there are multiple different sets of parallel resistors in this diagram so be careful with the math. Also, if you were to add them all together, you would be proceding down the path as mentioning in a), rather than b).
Lets go with b) instead. Firstly, can you see the multiple sets of parallel resistance? Which ones are they? What values do you get if you find the total resistance for each different set?
When you have these values, redraw the circuit with each parallel set's total resistance represented by a single resistor, and then find the voltage across each one using either Ohm's Law or the voltage divider equation.
Lastly, use your knowledge of how the voltage across each total resistance in your new diagram relates to each parallel resistor set in the original diagram, how voltages in parallel operate and Ohm's Law, and let us know the answers you get.
Zryn said:Yep, that's a good start.
Another convention to write combinations of resistors is using the '||' symbol for parallel combinations and the '+' symbol for series combinations.
You could therefore say that the total resistance of the circuit is Rt = R1||R2 + R3||R4||R5||R6.
... so I have now figured out that across R1 (R1 + R2) that it is 17.24Ohms ...
... and R2 (R3 + R4 + R5 + R6) is 18.32Ohms ...
... then used the voltage divider equation Vin * R1/R1+R2 to find that R1 = 5.82V and R2 = 6.18V ...
Now I'm confused about what to do next?
I'm going to take a stab at this and say =
80/300 * 18.32ohms = 4.89ohms (resistance segment of R4)
then using ohms law --> 6.18/4.89 = 1.26A ? Am I on the right track?
Zryn said:Given resistors in parallel, what do you know about the voltage across them and the current through them?
Isn't the voltage drop the same across all resistors in parallel? While the current is split
also sorry: Rt(R1) = 1/ (1/30 + 1/40)
Zryn said:Yep!
So given that you know the voltage across the resistors, you know that since they are in parallel that the same voltage will be across each individual resistor and you know the value of each individual resistance, you can now use Ohms Law to calculate the current through each one.
To check if you have the correct values as well, you can then add these currents up, and the result should equal the previously mentioned total circuit series current.
Alternatively, knowing the series current through the resistors and their resistance value you can use the current divider equation to calculate the current through each individual resistance.
To check if you have the correct values, you can multiply these currents by their associate resistances and you should get the previously mentioned parallel voltage.
That math looks good, when I put it into the calculator though I get 17.14R instead of your result of 17.24R. It's a very minor error, but as it propagates down through your equations it may result in the final answer being very wrong.
For this reason it's good to write out your math to show that you know what you are doing, and in the instance that you get the wrong number somewhere your markers will likely give you marks for knowing what you are doing, whereas were you to put down no math explaining your method, and then the final incorrect answer, you may get few marks at all.
Zryn said:Using the same method, can you find the current through R3, R5 & R6 as well?
Also, can you write out your algebraic equation for the current divider, like you did for the voltage divider earlier as Vin * R1/(R1+R2) ?
Ir4 = (Rt/R4) * It
Zryn said:The series current everywhere in the circuit is 0.337A, as calculated previously. When this current enters the parallel set of R3||R4||R5||R6 it will split according to the inverse resistance values (the path of least resistance will be the largest current) and then it will recombine when it exits the parallel set.
This means that the sum of the currents through these 4 resistors must be equal to the series current of the circuit.
Is = Ir3 + Ir4 + Ir5 + Ir6
0.337 = 0.133 + 0.15 + 0.171 + 0.200
0.337 = 0.654
Something is wrong.
Alternatively, you know that the voltage across each of these resistors is 6.2V. Therefore by Ohms Law:
V = Ir4 * R4
6.2 = 0.15 * 80
6.2 = 12
Something is definitely wrong.
The current divider equation is actually:
Ir = Rt / (Rt + Rn) * I
Where 'Rt' is the total resistance of the resistors which are *not* the resistor in question which you are trying to find the current through, 'Rn' is the resistor in question which you are trying to find the current through and 'I' is the source current.
Zryn said:Getting closer!
In this instance we are considering the parallel set of R3||R4||R5||R6 by it's lonesome. You could actually look at it as the following drawing (attached).
I'm actually confused now...
so it should be = 220/300 * 0.337
Where 'Rt' is the total resistance of the resistors which are *not* the resistor in question which you are trying to find the current through, 'Rn' is the resistor in question which you are trying to find the current through and 'I' is the source current.
Zryn said:Nope, current dividers are a little tricky. Reread the following and then look at the equation below to see what's happening:
Ir = Rt / (Rt + Rn) * I
Ir4 = (R3||R5||R6) / (R3||R5||R6 + R4) * I
Ir4 = ...
Zryn said:How do you combine resistors in parallel?
R3 + R5 + R6 as a series combination = 220R
R3||R5||R6 as a parallel combination = ... ?
(R3*R5*R6 / R3+R5+R6) / (R3*R4*R5*R6 / R3+R4+R5+R6) * 0.337 = 0.006A
Ir4 = Rt / (Rt + Rn) * I
Where 'Rt' is the total resistance of the resistors which are *not* the resistor in question which you are trying to find the current through, 'Rn' is the resistor in question which you are trying to find the current through and 'I' is the source current.
Zryn said:Much closer now .
Rt = (R3*R5*R6 / R3+R5+R6) = the total resistance of the resistors which are *not* the resistor in question which you are trying to find the current through ... Correct.
However, Rn = R4 = the resistor in question which you are trying to find the current through.
Therefore Rt + Rn = [ (R3*R5*R6 / R3+R5+R6) + R4 ] and not (R3*R4*R5*R6 / R3+R4+R5+R6).
Zryn said:My apologies, when I read what you said in post #29 I mistakenly thought I read something different to what was actually there, and I have since amended my post #30. Read it again and then comment, but you're on the correct path, just need to get the numbers right.
Zryn said:Yes you can, that is the correct answer and the discrepancy is only due to rounding errors.
The current divider that you were going to end up with is as follows:
Ir = Rt / (Rt + Rn) * I
Ir4 = (R3||R5||R6) / (R3||R5||R6 + R4) * I
Ir4 = [ 1 / (1/R3 + 1/R5 + 1/R6) ] / { [ 1 / (1/R3 + 1/R5 + 1/R6) ] + R4 } * I
Ir4 = [ 1 / (1/90 + 1/70 + 1/60) ] / { [ 1 / (1/90 + 1/70 + 1/60) ] + 80 } * 0.337
Ir4 = 23.77 / ( 23.77 + 80 ) * 0.337
Ir4 = 0.229 * 0.337
Ir4 = 0.077