# In this circuit, how can I find the value of voltage if I only have other voltages and resistances?

We just learned about Kirchhoff's laws, but no matter how much substitution I do I can't get more than the value of one current. Relevance

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

• 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.

• Anonymous
1 year ago

Formula.

• 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

• Mr. Un-couth
Lv 7
1 year agoReport

I agree. Sometimes the first solution one sees is not the simplest.

• 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 <-----

• 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.

• Just FYI, as an engineer the usual way to solve this would be Delta-Star 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).

• 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. • Anonymous
1 year ago

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 • Dan1 year agoReport

Yeah Linear circuits and systems

• 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.

• Dan1 year agoReport

I'm googling superposition and how to use it, but it's a chapter ahead of where we are currently in the textbook. I'm trying it now but I know there should be another way to solve this. Do you see a way to solve it with KVL or KCL? I just need v0, I was assuming I needed to find current to get there