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Finding Current and Voltage in a series parallel combination

Epircus

Feb 15, 2013
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I'm new to electronics and I'm really struggling to understand combination of series and parallel circuits.

I built this circuit recently and gathered all the required data across each resistor.(missing is the dc voltage source at 15V)
http://imgur.com/HZHptfV

I calculated Rt = 316 ohms and then using Ohms law found Is = 47.4mA. I found all this to be correct but that's not what I'm having difficulties with.

Now I want to be able to figure out how much the current is going through each resistor. I've been trying different formulas but have found no luck(Ix=rt/rx) for example I7 measured current is 7.40 mA using my formula it is way off. I know I'm doing something wrong and I want to have a better understanding of this circuit. Is there anyone willing to walk me through finding each branches current? I would highly appreciate the help.
 

john monks

Mar 9, 2012
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I mustn't do your home for you but I will suggest one thing. You already have the circuit and a 15 volt supply. Hook it up and measure the voltage of every resistor one by one. Find out the voltage and see if you can logically tell why each voltage is as it is. The math is easy so I assume you are having difficulties understanding resistors. Remember these things:
The current across a resistor is directly proportional to the voltage across it.
The resistance is directly proportional to the voltage divided by the current.
The voltage is directly proportional to the resistance times the current.
The current leaving a device is equal to the current entering the device.
Draw your schematic larger, draw current paths, and the answers should become clearer.
 
Last edited:

Laplace

Apr 4, 2010
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The usual method for calculating the performance of resistor networks is to substitute an equivalent resistance, Req, for the network. Do this by combining individual resistors. Combine parallel resistors into one resistor. Combine series resistors into one resistor. Keep doing this until you have a single resistor, Req. Then backoff one so you have R1 in series with Req-1 (the equivalent resistance that is one step short of including R1). Now you will have a voltage divider and can calculate the voltage across R2 in parallel with Req-2. Then go to the voltage divider with R3 and Req-3 and so forth until you have all the voltages across all the resistors. At that point you can then find all the currents.
 

Epircus

Feb 15, 2013
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Feb 15, 2013
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Thank you for your help so far.

Alright, so I sort of understand the voltage divider rule. And I am able to find V1.

V1 = 91*15v/91+225 = 4.32V

Thats equal to what I measured and with that I can now find the current going through the resistor.

I1 = 4.32/91 = 47.473 mA.

Up to this point I understand what I'm doing, going forward is where I'm getting confused. I came to this forum because I had my professor help me get to the next part and I'm pretty sure he is wrong in what he told me to do. Because what he has doesn't match what is measured.

Now v2 = 330*15/330+712 or V2= r2* E/ r2+req3

This is where I'm getting confused what am I doing wrong here? I'm not even getting closed to what is measured.
 

john monks

Mar 9, 2012
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From your math I quickly understand that your problem is not with the math but with your understanding of electronics. I have two items.
1. If your circuit is being powered by 15 volts and you know what the voltage across R1 is then what is the voltage across R2?
2. When using mathematical equations you must use units otherwise your equations are undefined. In this case your units are volts, milliamps, and ohms.

If you have trouble with any of this I am happy to go through this in full detail.
 

Epircus

Feb 15, 2013
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Feb 15, 2013
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Ah, ya, I'll admit I don't label well. I need to improve.

But I think I understand where you are coming from now.
Vab = Va-b
Va= 15v
vb = 4.3v
Vab = 10.7v

so r2 = 10.7v, correct? This matches what is measured.

So now I want to find r3 and am confused again. I have to leave for work, so I'll take a look at this on my break to see if I can't figure it out. Thanks for your insight so far!
 

Laplace

Apr 4, 2010
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Attached files illustrate two methods for calculating voltages in a resistor network. The pdf shows successive calculation of equivalent resistance for portions of the network so that voltages may be found from a resistor divider arrangement. The png shows the node equations from which the node voltages are solved directly. It's arithmetic versus algebra.
 

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Epircus

Feb 15, 2013
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Feb 15, 2013
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Thank you, I was actually able to figure it out with your guys help and understand it now.
 
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