Yahoo Answers: Answers and Comments for In this circuit, how can I find the value of voltage if I only have other voltages and resistances? [Engineering]
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From Dan
enUS
Sat, 08 Sep 2018 05:00:33 +0000
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Yahoo Answers: Answers and Comments for In this circuit, how can I find the value of voltage if I only have other voltages and resistances? [Engineering]
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From Mr. Uncouth: Actually the two Voltage sources are connected...
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Sun, 09 Sep 2018 05:00:52 +0000
Actually the two Voltage sources are connected series aiding. That means that the total Voltage equals 15 Volts. The 15 Volts is connected in series with two paralleled branches. The branch that you are concerned with contains a 100 Ohm resistor and a 200 Ohm resistor connected in series with the 15 Volt total.
Therefore: The current through the 100 Ohm resistor in this branch = V/R = 15V/(100 + 200)Ohm = .05 Amps
The Voltage drop across the 100 Ohm resistor = (Amps)*(Ohms) = (.05 A)*(100 Ohms) = 5 Volts
With Voltage polarity being observed the sum of the Voltages around any closed loop must equal zero Volts.
Therefore the sum of the Voltages around the lower right loop that contains Vo, the 100 Ohm resistor and the 5 Volt source must equal zero Volts.
thus; {(5 Volt source) + [5 Volts (across the 100 Ohm resistor)] + Vo} must = 0 Volts
Solve for Vo and get;
Vo = 5 Volts + 5 Volts = 0 Volts

From Roger: Call the left 100 ohm resistor R1, and the rig...
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Sat, 08 Sep 2018 23:02:21 +0000
Call the left 100 ohm resistor R1, and the right 100 ohm resistor R2
The 200 ohm resistor and R2 are in series, there total resistance is 300 ohm
Call the current that flows through them i1 and assume it flows clockwise
 10 Volts +i1(200 ohms +100 ohms)  5 Volt = 0 In this equation voltage rises are 
and voltage drops are +
You can rearrange the equation so voltage rises are on the right and voltage drops on
the left. I combined the two voltage rises i.e. Voltage drops = Voltage rises
and got this:
i1(300 ohms) = 15 volts
i1 = 0.05 amps
Vo= 5 Volts +i1(100 ohms) =  5 volts +(0.05)(100 ohm) = 0 Volts
Remember the current enters R2 on the right side so the voltage drop in the resistor
is positive and voltage rise from the 5 volt source is negative
Note we didn't need to solve for the current in R1
But it is easily done (call it i2 and have flow clockwise)
its current is found by
i2(R1) = 10V + 5V
i2 =15/100 = 0.15 amps

From billrussell42: I'd use superposition.
Short out 5v (red...
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Sat, 08 Sep 2018 11:00:23 +0000
I'd use superposition.
Short out 5v (red), you have a voltage divider from 10v to 200 Ω to Vo and 100 Ω to ground, so Vo = (100/300)10 = 3.33 v
Short out 10v (blue), you have a voltage divider from –5v to 100 Ω to Vo and 200 Ω to ground, so Vo = (200/300)5 = –3.33 v
Vo is the sum, zero volts.

From Anonymous: Is this university physics?
It is 5 volts, y...
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Sat, 08 Sep 2018 07:09:32 +0000
Is this university physics?
It is 5 volts, you ignore the variable V0 because it is basically the voltage from the bottom essential node to the right essential node. Essential node is basically a node that connect 3 or more wires if you are wondering. So you find the voltage through the RIGHT 100 Ohm resistor by doing a KVL on the right loop (triangle loop). and so from right to bottom is 5 volts after calculations.
I need to write something so my answer shows
Wait is is either 5 or 0 volts

From qrk: Use superposition, where you short out each so...
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Sat, 08 Sep 2018 05:12:31 +0000
Use superposition, where you short out each source independently and sum the individual voltage results at the output. The answer is surprising.
With that knowledge, you can figure out the current through each resistor.

From Who: junction 100 ohm resistors = 5 relative to 10...
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Wed, 19 Sep 2018 19:36:13 +0000
junction 100 ohm resistors = 5 relative to 10v ve and 5v +ve
so voltage drop to the junction = 15v
200 ohm in series with 100 ohm to this junction = so this gives you the current through 200ohm and therefore voltage drop across it relative to 10v +ve  (= 50 mA = 10V drop)
so Vo+ = 0v relative to 10v ve=
Vo = 0v

From Steven: You are suppose to spot the relationship betwe...
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Wed, 12 Sep 2018 15:24:56 +0000
You are suppose to spot the relationship between 10V/200=5V/100, so that Vo=0.
Of course the left 100 resistor is irrelevant since it connects between to defined voltages.

From Anonymous: Formula.
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Mon, 10 Sep 2018 12:09:38 +0000
Formula.

From oldschool: Using KVL assume clockwise i1 in the top loop ...
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Sun, 09 Sep 2018 03:24:54 +0000
Using KVL assume clockwise i1 in the top loop and clockwise i2 in the lower left. For the top loop
400i1 + 100i2 = 0 > i2 = 4i1
Now the lower left loop
100i1  100i2 = 15
Now substitute i2 = 4i1
100i1 400i1 = 15 > 15 = 300i1 > 0.05A = i1
So i2 = 0.2A
Vo = +100i1  5 = 5  5 = 0 <

From Philomel: Assume that the left 100Ω resistor is R1, the ...
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Sun, 09 Sep 2018 00:49:52 +0000
Assume that the left 100Ω resistor is R1, the 200Ω is R2 and the right 100Ω is R3.
Circuit simulator applet says :
R1 15v 150mA L>R
R2 10v 50mA L>R
R3 5v 50mA R>L
If you trace from the top of Vo through R3 through the 5v source you will have a drop of 5v then a rise of 5v equalling 0V on Vo.
This is why I use the Circuit Simulator App. It always works. When in doubt build the circuit and test reality not theory.
Vo = 0V.

From Dixon: Just FYI, as an engineer the usual way to solv...
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Sat, 08 Sep 2018 12:27:46 +0000
Just FYI, as an engineer the usual way to solve this would be DeltaStar transformation on the three resistors.
This is probably not something they teach at school or in physics but it isn't difficult. It basically amounts to replacing components in a delta or star formation with different value components in the other formation that have the same net behavior when viewed from the three nodes, ie an equivalent circuit. This then allows simplification of the whole circuit by normal series / parallel reduction. The reason we do this preferentially is that it doesn't involve any thinking! The transformation is formulaic and super quick if you drop the values into a program or app and reduction of series and parallel circuits is second nature.
In fact my experience is that engineers never set up simultaneous Kirchhoff equations for circuit analysis, we do it all with transformations (including swapping between Thevinin and Norton equivalents).