- #1

joaofbi1

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Here's the question:

Thank you.

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- Thread starter joaofbi1
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- #1

joaofbi1

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Here's the question:

Thank you.

- #2

NascentOxygen

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Hi joaofbi1. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Let's see your thoughts on how to solve this. Then we can see where you may be going wrong.

First step, draw in some current paths so you can see where the current is flowing.

Let's see your thoughts on how to solve this. Then we can see where you may be going wrong.

First step, draw in some current paths so you can see where the current is flowing.

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- #3

UltrafastPED

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6V = 2.2k(I1-I2)

0 = 2.2k(I2-I1) + 10k(I2) + 1k(I2-I3)

0 = 1k(I3-I2) + 3.3k(I3)

Then solve the three simultaneous linear equations via your favorite method (substitution, elimination, Cramer's rule), and I3 is your answer.

I also considered a Delta-Y transform but it didn't seem to make the solution go any faster.

- #4

joaofbi1

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Rt= R1XR2 / R1 + R2 so it would've been Rt = 1x3.3 / 1 + 3.3 which gave me the answer of 0.8kΩ

After that I noticed that 2.2kΩ, 10kΩ and 0.8kΩ was kind of in a circuit, if you rearranged it, because of that I used the formula:

Rt = R1 + R2 + R3 which would have been Rt= 2.2 + 10 + 0.8 which gave me 13kΩ

After that I used the formula I= V/R - But before that I converted 13kΩ into Ω by ten to the power of -3 so I got 13x10(-3).

After that I did Voltage / Resistance -> 6/(13x10(-3)) which gave me the current of 462Amps which then I converted into mA and got 0.462mA.

That's where I got to, now I'm stuck.

- #5

joaofbi1

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6V = 2.2k(I1-I2)

0 = 2.2k(I2-I1) + 10k(I2) + 1k(I2-I3)

0 = 1k(I3-I2) + 3.3k(I3)

Then solve the three simultaneous linear equations via your favorite method (substitution, elimination, Cramer's rule), and I3 is your answer.

I also considered a Delta-Y transform but it didn't seem to make the solution go any faster.

I've just started the electronic course so I'm quite new to the method that you're using. So I don't quite understand it.

- #6

UltrafastPED

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Also you cannot sum the resistors around the middle loop ... this is only valid for resistors in series; resistors in series appear 1-2-3 along the same piece of conductor without any additional branches. In this case you have an additional branch which has the 6V voltage source.

It is possible to continue the reduction of the resistor network into a single loop with the voltage source plus one resistor in series. This would provide the total current being supplied by the battery, but now you have lost track of the 3.3k resistor!

The easiest way is to apply the Kirchoff voltage law (KVL) around each loop, or the Kirchoff current law (KCL) at each node, or apply a method (like the mesh method) which is derived from those laws.

- #7

UltrafastPED

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You know Ohm's law; the next step is usually to teach KVL and KCL followed by some practice.

The mesh method is simply a faster, systematic application of KVL; nodal analysis is a faster, systematic application of KCL. In this case I used the mesh method; you can arrive at the same results via KVL or KCL though the equations may have a different appearance ... but the three currents will be the same.

- #8

joaofbi1

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Is this not possible to do with Ohm's law? If it is then could you help me out using Ohm's law please?

- #9

UltrafastPED

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If you now redraw the circuit with the reduction (using the 0.8 kOhm resistor) you should see that there is now a series combination of two resistors which can be combined; do this, and redraw the circuit.

This time you should see two resistors in parallel with the battery. You can combine the two resistors and use Ohm's law to find the total current.

Do that, keeping all of the circuit diagrams ... then you can start solving backwards, but first things first!

- #10

joaofbi1

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then it turns into this:

Right? Then what shall I do, this is the bit where I get confused on what to do with the two resistors.

- #11

NascentOxygen

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Keep it as 3 resistors, 2.2, 10 and 0.8. Draw some current paths, these being shown as closed loops.

- #12

joaofbi1

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Keep it as 3 resistors, 2.2, 10 and 0.8. Draw some current paths, these being shown as closed loops.

What do you mean by draw circuit paths?

- #13

UltrafastPED

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So combine the 10k with the 0.8k in series: 10.8k, and then you are left with two resistors in parallel, the 2.2k and the 10.8k. Combine them and then solve for the total current.

- #14

joaofbi1

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So combine the 10k with the 0.8k in series: 10.8k, and then you are left with two resistors in parallel, the 2.2k and the 10.8k. Combine them and then solve for the total current.

Oh I understand what you mean now, so once I've found the total current what formula do I use to find the current going through the resistor?

- #15

UltrafastPED

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For the first step you will have two resistors in parallel; part of the total current will pass through each branch.

Since one of the resistors is the 2.2k resistor from the original circuit, you can go back and label both the total current drawn from the battery, and the current passing through the 2.2k resistor. The remaining current must pass through the 10k resistor.

What will then happen to this part of the current?

- #16

joaofbi1

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- #17

UltrafastPED

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You have two resistors in parallel on this diagram ... the next to last circuit ... one is 2.2k, the other is 10.8k.

They both see the same voltage: 6 volts. So you can calculate the current through each of these two paths via Ohm's law. You need to do this so that you know the current going through the 10.8k equivalent resistor.

You will use that current for the next step.

I get 1.98k (2k) for the total equivalent resistance, which means that the total current is

I=6v/2k Ohms=3 mAmps. You should always keep track of your units.

Correction: I get 1.83 kOhms for the equivalent resistance, and 3.3 mAmps for the total current - so your total current is correct.

They both see the same voltage: 6 volts. So you can calculate the current through each of these two paths via Ohm's law. You need to do this so that you know the current going through the 10.8k equivalent resistor.

You will use that current for the next step.

I get 1.98k (2k) for the total equivalent resistance, which means that the total current is

I=6v/2k Ohms=3 mAmps. You should always keep track of your units.

Correction: I get 1.83 kOhms for the equivalent resistance, and 3.3 mAmps for the total current - so your total current is correct.

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- #18

joaofbi1

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- #19

UltrafastPED

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The current through the 10.8 kOhm resistor will then be the same as the current through the 10 kOhm resistor in the original circuit: it will then divide as it goes through the original two parallel resistors, the 1k and the 3.3k ohm resistors.

You have two ways to solve this final step.

- #20

joaofbi1

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So once I get the current for the 10.8 resistor do I divide that current by 1and 3.3

- #21

NascentOxygen

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Not exactly.So once I get the current for the 10.8 resistor do I divide that current by 1and 3.3

But you could determine how much of the battery's voltage is "lost" because of the 10kΩ resistor, so the remainder is what appears across the pair of parallel resistors.

- #22

joaofbi1

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Im quite confused, what calculations would I need to perform?

- #23

NascentOxygen

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How many volts are lost across the 10k resistor? Use Ohms Law.Im quite confused, what calculations would I need to perform?

- #24

joaofbi1

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- #25

joaofbi1

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- #26

UltrafastPED

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It should be 3.3 mA through 10k ohms. Pay attention to your units!

And that is the total current; we previously calculated the current for this branch - what is it?

And that is the total current; we previously calculated the current for this branch - what is it?

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- #27

joaofbi1

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So how do I calculate how much mA I've lost through the resistor? What formula?

- #28

UltrafastPED

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I1 = 6 V/2.2k ohms = 2.73 mA; I2 = 6 V/10.8 k ohms = 0.55 mA. These add up to 3.3 mA total current which we obtained from the final circuit.

So the current through the 10 k ohm resistor will be I2 = 0.55 mA.

This is the current that will be available to the parallel resistors 1 k and 3.3 k ohm; when they were combined they provided the addition 0.8 k ohms that was combined with the 10 k resistor in series to obtain the 10.8 k ohm equivalent resistor.

This process of working backwards, and marking the resulting currents on each as we walked backwards up the circuits has provided all of the information needed.

- #29

joaofbi1

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It should be 3.3 mA through 10k ohms. Pay attention to your units!

And that is the total current; we previously calculated the current for this branch - what is it?

So how do I Calculate how much mA I have lost through the resistor?

- #30

UltrafastPED

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Or we could have solved from the equivalent resistance of the parallel resistors which was 770 Ohms, with a current of 0.55 mA gives V = 0.770 kOhms x 0.55 mA = 0.44 [V]

These voltages are close (0.5, 0.44) but we should get the same answer both ways ... this indicates that there is an error earlier in the calculations which you should recheck.

Assuming that 0.5 V is correct you can use Ohm's law to get the current I3 = 0.5 V / 3.3 k Ohms = 0.15 mA.

- #31

UltrafastPED

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So how do I Calculate how much mA I have lost through the resistor?

No current is lost through a resistor; only the voltage is used up.

- #32

joaofbi1

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But where exactly did you get 0.55mA from? That has completely stumped me.

- #33

UltrafastPED

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Current I2 from message #28.

- #34

joaofbi1

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Could I not just simply do 6V/10kΩ to get 0.6 instead of doing

I2 = 6 V/10.8 k ohms = 0.55 mA

- #35

joaofbi1

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Shall I convert the 0.55mA into volts then do the same operation as above?

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