Can someone explain how you reach the value that's shown on the meter?

Thank you,

- Nicky

 

Looks like homework. Smells like homework. Moved to homework.

 

Do you know how to do superposition?

 

Do you know how to do superposition?

Do you know those cheesy vampire movies where as soon as vampire shows up, the protagonist pulls out a wooden cross? Yeah, I think we will use Superposition for the cross.

 

Using superposition in this situation kind of confuses me? I actually made that diagram using Proteus, where the voltage meter is, is suppose to be an open circuit for thevenin's theorem, maybe I messed up by putting in the voltage meter?

 

Initially, just forget that the meter is there. What is the circuit? Two batteries & two resistors. Can you find the voltage drop across each resistor?

 

What I do know is

I = (25 + 35) / (9 + 4)

I can work out potential difference across the resistors but I'm confused what I'm subtracting it from and the direction of stuff. I assumed the 25V battery is going in the direction of the 9ohm resistor but as there isn't any voltage drop at the junction of the open circuit, I always think that shouldn't all 25 volts go down there? Then the same for 35V as it goes the other way...And as they appose each other it would give a net voltage of 10V...

As you can see I'm a mess ;x help

 

as they appose each other it would give a net voltage of 10V.

If the batteries oppose each other, then why do you add them together to calculate loop current with 60 volts?

 

Here's how you solve it with superposition.
The small plate of the batteries are positive.
Calculate the voltage across the meter with Bat2 replaced by a short circuit.
Then replace Bat2 and calculate the voltage across the meter with Bat1 replaced with a short circuit.
(If it helps your calculations, you can ground the minus side of the meter.)

Add the two voltages (observing the polarity) and you will have the correct answer.

 

Here's how you solve it with superposition.

But doing superposition means calculating the loop voltage drops twice in order to add the voltages. Better to add the battery voltages once, then calculate the loop voltage drops once.

Superposition violates the Law of Wally.

 

But doing superposition means calculating the loop voltage drops twice in order to add the voltages. Better to add the battery voltages once, then calculate the loop voltage drops once.

Superposition violates the Law of Wally.

It's easier for me to calculate the voltages drops twice and subtract once as compared to calculating the voltages drops once and figuring out how to add/subtract the battery voltages to get the correct answer (which is what the OP is having a problem with).

That's this Wally's law.

But to each his own.

 

Here's how you solve it with superposition.
The small plate of the batteries are positive.
Calculate the voltage across the meter with Bat2 replaced by a short circuit.
Then replace Bat2 and calculate the voltage across the meter with Bat1 replaced with a short circuit.
(If it helps your calculations, you can ground the minus side of the meter.)

Add the two voltages (observing the polarity) and you will have the correct answer.

This seems to be contrary to convention: https://en.wikipedia.org/wiki/Electronic_symbol

and would result in a reading on the meter of -6.54 volts.

 

This seems to be contrary to convention:

Yup, I got it backwards.

 

What I do know is

I = (25 + 35) / (9 + 4)

I can work out potential difference across the resistors but I'm confused what I'm subtracting it from and the direction of stuff. I assumed the 25V battery is going in the direction of the 9ohm resistor but as there isn't any voltage drop at the junction of the open circuit, I always think that shouldn't all 25 volts go down there? Then the same for 35V as it goes the other way...And as they appose each other it would give a net voltage of 10V...

As you can see I'm a mess ;x help

Get in the habit of properly tracking your units. You should have

I = (25 V + 35 V) / (9R + 4R) = (60 V) / (13R)

Now annotate the diagram with the this current showing the direction in which it is flowing. Ignore the meter -- physically delete it from the drawing if you need to.

Now start at the node that the negative terminal of the meter is (or was) connected to. Call that node 0 V (you get to do that for exactly one node in the circuit).

Then toss a coin and choose to either follow the circuit clockwise or counter clockwise from there.

The just walk one component at a time along that path and calculate the voltage on the other side of the component starting with the voltage on the near side and taking into account the voltage drop/gain across the component.

Take you best shot at doing this and post the results (including the appropriate annotated diagram, if possible).

 

Get in the habit of properly tracking your units. You should have

I = (25 V + 35 V) / (9R + 4R) = (60 V) / (13R)
...

The resistors notation is confusing but from the result shown by volt meter, the resistances of R1 and R3 should be 4 ohms and 9 ohms, respectively not 4R and 9R as symbolic values.

 

The resistors notation is confusing

Why ??
For R you can pick any real number you like.
For R = 2.2kΩ we have
4R = 4*2.2kΩ = 8.8kΩ
9R = 9*2.2kΩ = 19.8kΩ

the resistances of R1 and R3 should be 4 ohms and 9 ohms, respectively not 4R and 9R as symbolic values.

For this case (4Ω and 9Ω) we have R = 1Ω

 

The resistors notation is confusing but from the result shown by volt meter, the resistances of R1 and R3 should be 4 ohms and 9 ohms, respectively not 4R and 9R as symbolic values.

That's possible. A common notation on schematics (particularly in days past) was to use the units multiplier (m. k. M. etc) as the decimal point and, if there was no multiplier, use R (for a resistor) instead.

In this case it doesn't really matter since you will get the same results regardless of what you choose for R.

 

Why ??
For R you can pick any real number you like.
For R = 2.2kΩ we have
4R = 4*2.2kΩ = 8.8kΩ
9R = 9*2.2kΩ = 19.8kΩ


For this case (4Ω and 9Ω) we have R = 1Ω

That's possible. A common notation on schematics (particularly in days past) was to use the units multiplier (m. k. M. etc) as the decimal point and, if there was no multiplier, use R (for a resistor) instead.

I didn't know that!

In this case it doesn't really matter since you will get the same results regardless of what you choose for R.

Oh, yes, my mistake. You are right. The result remains the same no matter which value of R.