# Voltage and Circuits

I need someone to explain how to get the answer for (c)
I know its like this (from solution manual)...

Va = Vc + 8 + I(1.4 + 5)

From what it seems, they are just adding voltage in the counter clockwise direction. How are they doing this?

How come I can't add voltage from the clockwise direction?

I am very confused to how this works, I can't seem to grasp an image of what is going in circuits (especially with 2 batteries!) Please explain everything to me.

Last edited:

chroot
Staff Emeritus
Gold Member
You can add voltages in either direction you like! You can pick a specific point, say point a, and call that "ground." You can simply declare that point a has zero volts of potential. Then walk the loop, adding the voltages as you go. The sum of all those voltage drops has to equal zero.

- Warren

You can add voltages in either direction you like! You can pick a specific point, say point a, and call that "ground." You can simply declare that point a has zero volts of potential. Then walk the loop, adding the voltages as you go. The sum of all those voltage drops has to equal zero.

- Warren

So lets say I decided to add them up in the clockwise direction.

Va = 15.248 + (-I*9) + Vc , Would this work also? I tried this and answer is a little bit different.

Current is not a vector so what determines if voltage is negative?

chroot
Staff Emeritus
Gold Member
Let's write this out a bit more explicitly! I'm going to start with point a, call it zero volts. I'm also going to assume that the current is flowing in the clockwise direction, since that makes it easier for me in my own head.

First, we encounter a resistor. The voltage drop across a resistor is v = R*i. Next, we encounter a battery. The battery maintains a 16V difference between its positive and negative terminals, so, moving clockwise, the voltage drops 16V. Continuing this around this loop, I get:

0 -1.6*i - 16 - 9*i + 8 - 1.4*i - 5*i = 0

Does this make sense so far?

- Warren

Let's write this out a bit more explicitly! I'm going to start with point a, call it zero volts. I'm also going to assume that the current is flowing in the clockwise direction, since that makes it easier for me in my own head.

First, we encounter a resistor. The voltage drop across a resistor is v = R*i. Next, we encounter a battery. The battery maintains a 16V difference between its positive and negative terminals, so, moving clockwise, the voltage drops 16V. Continuing this around this loop, I get:

0 -1.6*i - 16 - 9*i + 8 - 1.4*i - 5*i = 0

Does this make sense so far?

- Warren
A little bit. Ok so another example..
Lets say the polarity of the 16v battery is switched. So now, the + side is on the right and the negative side is on the left. ALSO, the 1.6 ohms is on the right side of the battery.

Now I want to go from Va to Vb. Would this be correct?

Va = 16 -1.6I + Vb so
Va - Vb = 16 - 1.6I

Then I want to go from Vb to Va so..
Vb = 1.6*-I - 16V + Va

Does these 2 equation look correct??

chroot
Staff Emeritus
Gold Member
I don't think it's a good idea to change the problem around in the middle of trying to solve it, especially when trying to get help on a forum. That said, the equation should be:

Va + 16 - 1.6i = Vb

In your head, say it like this: "start at point a (with potential Va), add 16 volts, subtract 1.6*i volts, and now arrive at point b (with potential Vb)."

You're trying to start with Va and set it equal to things -- don't do that! You're likely to make sign mistakes. Instead, start with a known voltage (like Va), and add things directly to it as you walk around the loop.

- Warren

I don't think it's a good idea to change the problem around in the middle of trying to solve it, especially when trying to get help on a forum. That said, the equation should be:

Va + 16 - 1.6i = Vb

In your head, say it like this: "start at point a (with potential Va), add 16 volts, subtract 1.6*i volts, and now arrive at point b (with potential Vb)."

You're trying to start with Va and set it equal to things -- don't do that! You're likely to make sign mistakes. Instead, start with a known voltage (like Va), and add things directly to it as you walk around the loop.

- Warren
This is another problem in the book. I switched it so I can see if I understand.
So from what you said, lets start at point Vb now. We are still using negative on the left and positive on the right.

So..
Vb - 1.6I - 16V = Va ?

In my head.." start at Vb, subtract 1.6I because its a voltage drop. Then subtract another 16V because its another voltage drop. "
Correct? x_x

chroot
Staff Emeritus
Gold Member
Ok, I don't know what you're doing now...

You're back to the original problem, right? And now you're starting at Vb, and walking the loop counter-clockwise, assuming that the current flows counter-clockwise?

- Warren

No, I said we are still using the configuration of - on the left of the battery and + on the right of the battery and the 1.6 ohm is on the right of the battery. I am just going counter-clockwise now from Vb to Va. Sorry for the confusion.

chroot
Staff Emeritus
Gold Member
Yes, I think you're correct, then. Like I said, it's a really bad idea to change the problem in the middle of a post and expect people to keep up with you.

Can we get back to the original problem?

- Warren

Yes, I think you're correct, then. Like I said, it's a really bad idea to change the problem in the middle of a post and expect people to keep up with you.

Can we get back to the original problem?

- Warren

Ok back to the original problem. Va to Vc (Going counter clockwise)

Va - 5I - 1.4I - 8V = Vc
Va - Vc = 8V + 5I + 1.4I

And going from Vc to Va (This is for my understanding, going clockwise)

Vc + 8V - 1.4I - 5I = Va
Vc - Va = -8V +1.4I + 5I

NOW... We go Clockwise:
Va to Vc
Va - 1.6I - 16V - 9I = Vc
Va - Vc = 1.6I + 16V + 9I

Then Vc to Va (Counter Clockwise)
Vc - 9I + 16V - 1.6I = Va
Vc-Va = 9I - 16V + 1.6I

I hope this is not confusing. I tried my best to be clear

chroot
Staff Emeritus
Gold Member
I think you're on the right track; can you answer the questions now?

- Warren

Yea the answer is 11V I already know this from the solution. I will check this result with all my equation and see if it is right. Thanks alot!

Ok so what I figured out is that if you are going AGAINST the current, you need to ADD the IR

So for this equation

NOW... We go Clockwise:
Va to Vc
Va - 1.6I - 16V - 9I = Vc
Va - Vc = 1.6I + 16V + 9I

The correct way is

NOW... We go Clockwise:
Va to Vc
Va + 1.6I - 16V - 9I = Vc
Va - Vc = -1.6I + 16V - 9I