Why Does a Parallel Circuit Show Different Voltage Readings?

[rock42]

Homework Statement

https://photos-1.dropbox.com/pi/xl/rLbkcUie-3Lbcrb5YgTa73zJKs4agFowWz6uGFoBq7s/38240/1324436400/1241536/

the image seems to be broken, so here is a link
http://db.tt/fMwm8TrL

I recently picked up a future textbook of mine for something to do over the holidays, so forgive me for asking such a low level question, but I was hoping someone could explain why (in part a) the voltage should be -2V? I understand that, given the answer to part a, that b would be 16 W delivered. Thanks.

 

[phinds]

So, are you under the impression that folks here are psychic and can divine what you are talking about from the little you have said. No diagrams? No statement of the problem? No hint of what this is all about?

Wait ... my psychic power is kicking in ... I've GOT it. It's about electricity !

EDIT: OK, I see that you have edited you post to include an image, but I can't see it for some reason.

 

[gneill]

Homework Statement

https://photos-1.dropbox.com/pi/xl/rLbkcUie-3Lbcrb5YgTa73zJKs4agFowWz6uGFoBq7s/38240/1324436400/1241536/

I recently picked up a future textbook of mine for something to do over the holidays, so forgive me for asking such a low level question, but I was hoping someone could explain why (in part a) the voltage should be -2V? I understand that, given the answer to part a, that b would be 16 W delivered. Thanks.

What is the value of i_b? And so, what is the value of the controlled voltage source?

 

[NascentOxygen]

explain why (in part a) the voltage should be -2V?

Imagine that the voltage source in the centre was not present. What would be the terminal voltage of the other two elements?

So the question involving mention of the "interconnection" means what voltage v_g could you apply yet not cause any upset to the circuit operating without that v_g source being present. (It's a strange way of wording a question.)

 

[DeeNos]

I appreciate your curiosity and desire to understand the concepts in your textbook. In order to fully understand why the voltage should be -2V, we first need to understand the basic ideal circuit elements involved in this problem. These include resistors, which limit the flow of electric current, and voltage sources, which provide a potential difference for the current to flow through.

In this specific circuit, we have a voltage source of 8V and a resistor of 4Ω. According to Ohm's Law (V=IR), the voltage across a resistor is equal to the current flowing through it multiplied by the resistance. In this case, we can calculate the current as 8V/4Ω = 2A.

Now, in part a of the question, we have a second resistor of 2Ω connected in parallel with the first one. In a parallel circuit, the voltage across each branch is the same. So, if the voltage across the entire circuit is 8V, the voltage across the second resistor must also be 8V.

Using Ohm's Law again, we can calculate the current through the second resistor as 8V/2Ω = 4A. Since the total current in the circuit must equal the sum of the currents in each branch, the total current must be 2A + 4A = 6A.

Now, let's look at the voltage source. Voltage sources are defined as having a positive terminal and a negative terminal. In this circuit, the positive terminal is connected to the first resistor and the negative terminal is connected to the second resistor. This means that the voltage at the positive terminal is higher than the voltage at the negative terminal.

Since the voltage at the negative terminal is 8V, and the voltage at the positive terminal is 8V higher, the voltage at the positive terminal must be 16V. However, since we are dealing with a parallel circuit, the voltage across each branch is the same. This means that the voltage across the first resistor must also be 16V.

But wait, we calculated the voltage across the first resistor earlier as 8V. How can it be both 8V and 16V? This is where the concept of polarity comes in. Polarity refers to the direction of current flow in a circuit. In this case, we can see that the current is flowing from the positive terminal to the negative terminal, which means