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# Understanding Voltage Drop with Resistances.

## Understanding Voltage Drop with Resistances.

#1
10-24-14, 09:34 AM
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Join Date: Jan 2011
Posts: 50
Understanding Voltage Drop with Resistances.

Hi, I just started a course (Basic Electricity) and I was wondering if you could help!

DC = 60 V
R1 = 6.8 k ohms (in series)
R2 = 12 k ohms R3 = 6.2 k ohms R4 = 12k ohms (in parallel)

Note: R1 in series and the rest in parallel (R2, R3, R4).

1/R t = 6.2/74.4 + 12/74.4 + 6.2/74.4 = 24.4/74.4
R t = 1 / (24.4/74.4) = 1/0.327957 = 3.049 Ohms
Total R for parallel is 3.049 ohms
Total R for in series is 6.8 k ohms
then Total Resistance for both in series and parallel is = 9.849 ohms.

Total Current: I t = V/R = 60 V / 9.849 Ohms = 6.092 mA

V 6.8 = 6.8 ohms x 6.092 mA = 41.423 V

Calculate the drop in voltage:

V 111 = 60 V - 41.423 V = 18.577 V

Questions:

I can understand the calculation above but I do not understand where the voltage drop is. I have software where I drew the above example and used Multimeters (virtual) that I have created on each Resistance. The "+" probe goes before the resistance and the "-" probe goes after the resistance. Here are the readings I get:

R1 = 41.425 V
R2 = 18.575 V
R3 = 18.575 V
R4 = 18.575 V

If you look at the my calculations (above) it matches the results, i.e.

V 6.8 = 6.8 ohms x 6.092 mA = 41.423 V
V 111 = 60 V - 41.423 V = 18.577 V

What I'd like to understand is where is the voltage drop is, in the diagram ????
i.e. I do see the number on the multi-meters, but where and what are the Voltages on the wires, locations ???

Like where exactly the 41.423 V is ?
And where the 18.577 V is ?

Any help would be greatly appreciated,
M.B.

#2
10-24-14, 10:04 AM
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Join Date: Mar 2006
Location: USA
Posts: 29,711
Energy as described by amps x volts becomes heat and is dissipated to the surrounding air.

#3
10-24-14, 11:24 AM
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Join Date: Feb 2006
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Posts: 6,396
1/R t = 6.2/74.4 + 12/74.4 + 6.2/74.4 = 24.4/74.4
This formula is incorrect for the three parallel resistances.

#4
10-24-14, 11:45 AM
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Join Date: Oct 2013
Posts: 511
After R1, the three resistors in parallel receive the same voltage because they have a common connection point.
If all four resistors were in series, you would have a drop at each one and different voltages being read at each resistor.
And your three parallel resistors can be mathed out into one resistor, which you have the correct answer for but I don't understand your math for it. I was taught like so...
1/12 + 1/6.2 + 1/12 = 1/answer = 3.049 Ohms

#5
10-24-14, 11:45 AM
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Join Date: Mar 2010
Location: USA
Posts: 4,529
Voltage drop occurs everywhere in the circuit including in the wires, but for the purposes of this discussion, assume that all of the voltage drop occurs in the resistors.

The voltage drop in each of the resistors R2, R3, and R4 is the same. Measuring between where one end of R2 and R3 and R4 are tied together and where the other end of R2 and R3 and R4 are tied together you get X volts (voltage drop within that portion of the circuit). Measuring between one end of R1 and the other end of R1 you get Y volts (voltage drop). (X plus Y equals the supply voltage in this circuit.)

I don't recall the formula for calculating the net resistance of the parallel connected array of R2, R3, and R4 so I can't give you the exact numbers for Y and then X.

Meanwhile the current is the same within R1 and within the array R2, R3, and R4.

Last edited by AllanJ; 10-24-14 at 12:22 PM.
#6
10-24-14, 12:40 PM
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Join Date: Feb 2006
Location: USA
Posts: 6,396
Circuit

Meanwhile the current is the same within R1 and within the array R2, R3, and R4
The current in the parallel resistances is inversely proportional to the resistances.

The total resistance of the circuit is 6.8 ohms + 3.049 ohms = 9.849 ohms.
Total current is 60 V / 9.849 ohms = 6.09 amps.

The voltage drop at R1 is 6.8 ohms x 6.09 amps = 41.41 volts.

The voltage drop across each of the 3 parallel resistances is 60 volts - 41.41 volts =
18.59 volts.

#7
10-24-14, 02:31 PM
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Join Date: Oct 2008
Location: New England
Posts: 9,460
Hi papasmurf,
Not sure where you got your equation, but there are two ways to determine the value of parallel resistors. If you have just two, use the "product over sum" calculation for a quick easy solution. If you have more than two in parallel the equation is as Mr.Awesome posted, "the reciprocal of the sum of the reciprocals".

Then you determine the voltages as Wirepuller explained.

Many of us here enjoy basic electronics so feel free to bring any questions you have to the forum.

Bud

#8
10-24-14, 02:54 PM
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Location: port chester n y
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1/ RT = 1/R1 + 1/ R2 + 1 / R3 ---- 12K Ohms in parallel with 12K ohms = 6K Ohms = a resistance value exactly 1/2 of the value of each similar resistor.

6.8 / 9.85 = .69 of RT 3.05 / 9.85 = .31 of RT

VD across 6.8k resistor = 60v x .69 = 41.4v VD across 3.05k resistor = 100 x .31 = 18.6v

#9
10-25-14, 02:37 AM
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Join Date: Mar 2007
Location: Norway
Posts: 407
What I'd like to understand is where is the voltage drop is, in the diagram ????
i.e. I do see the number on the multi-meters, but where and what are the Voltages on the wires, locations ???

Like where exactly the 41.423 V is ?
And where the 18.577 V is ?

Understanding the voltagedrops just by reading this text may need a skill to visualize text.

The applied voltage (this could have been an rural telephone line, or even a fork lift battery) will drop to null during its travel in the circuit. The voltage will drop with equal part of the circuit represented by the resistors. Resistors in parallel will in this hence divide the current in different values, but in hence of voltage be seen as one resulting resistor.

If we looks pretty rough on your circuit, we start at on pole of the battery at 60V (=100 %), then we loose voltage in the first resistor of 6,8 k just lik pressure drop in a water pipe with a nearly closed valve. This are equal to 69% of the total resistance, and the voltage drop =69% of 60V. (slightly over 41V) Since the rest of the circuit are a resultant resistance of 3 resistors in parallel the rest of the voltage should be "lost" here.
And yes, thats right. 1/R=1/12+1/6,2+1/12 =approx 1/3,05 (k)

If you only knows the data given, you are stuck if you not are able to replace the 3 parallels with a resultant, and the simplify the circuit to one with 2 resistors in series. (of 6,8 and 3,05)

If you look further on this you will get the same voltage drop undependent of if the resistors are in ohms, kilo-ohoms or any other unit as long as they make the same percentage of the total resistance.

dsk

#10
10-25-14, 05:10 AM
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Join Date: Feb 2006
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Posts: 6,396
Question

What I'd like to understand is where is the voltage drop is, in the diagram ????
What diagram?..........................

#11
10-26-14, 10:49 AM
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Join Date: Feb 2002
Location: port chester n y
Posts: 1,983
Let "RP" = 1 / R1 + 1 / R2 + 1 / R3 + 1 / RN ---- then 1 / RT = RP----multiplying both sides of the equation by RT gives 1 = RT x RP---- dividing both side by RP gives RT = 1 / RP

#12
10-26-14, 05:29 PM
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Join Date: Jan 2011
Posts: 50
Thank you

Many thanks you all. I appreciated the reading. It did shed a light on a few things.

Sorry, but I couldn't upload the diagram at the very first thread. I got the popup window, browsed to my file, selected it and nothing.. I'm retrying it today and it looks liked it work... I guess I'll know for sure after I submit this thread.

I'm a newbie at this, I'm sure I'll have more questions down the road.

Thanks again,
MB (Papasmurf)