Finding the values of voltage and current in this circuit

[rugerts]

Homework Statement

It's best if one looks at the picture of the circuit diagram. What we're given is a current of 0.25 A and two resistances of 20 and 30 ohms.

Homework Equations

Kirchhoff's Voltage Law, Kirchhoff's Current Law, Ohm's Law

The Attempt at a Solution

My attempt at a solution started with applying KVL to the first mesh (the leftmost) and I arrived at a value of 5V. Although I am a little confused as to the sign conventions here. I've read that it's best to follow passive sign convention, which from my understanding, says that current going into a resistor is labeled with a + polarity. Here, I go around the loop clockwise but arrive at a negative voltage value, which to me suggests the drawn arrow in the given diagram is going opposite to its true direction.
To find current 2, I tried applying KCL, which for me resulted in a value of current 2 being the same as current 1, which is not correct. Can anyone shed some light on why I must use Ohm's Law instead of KCL here to find current 2?

 

[phinds]

You need to reconsider I2

How in the world did you conclude that -I1 + I2 = 0 ?

 

[rugerts]

You need to reconsider I2

How in the world did you conclude that -I1 + I2 = 0 ?

I was thinking that those two currents converge at node A like this:

 

[phinds]

Where did the zero come from?

 

[rugerts]

Where did the zero come from?

Doesn't Kirchhoff's Current Law state that the sum of currents at a node is 0? I believe I did the same thing on a previous problem and got a correct answer. See below:

 

[CWatters]

Homework Statement

It's best if one looks at the picture of the circuit diagram. What we're given is a current of 0.25 A and two resistances of 20 and 30 ohms.

Homework Equations

Kirchhoff's Voltage Law, Kirchhoff's Current Law, Ohm's Law

The Attempt at a Solution

My attempt at a solution started with applying KVL to the first mesh (the leftmost) and I arrived at a value of 5V.

Correct.

Although I am a little confused as to the sign conventions here. I've read that it's best to follow passive sign convention, which from my understanding, says that current going into a resistor is labeled with a + polarity. Here, I go around the loop clockwise but arrive at a negative voltage value, which to me suggests the drawn arrow in the given diagram is going opposite to its true direction.

The arrows on a circuit diagram do NOT tell you the direction that current is flowing. They tell you which direction the named current (eg I1, I2, I3, I4...) would be flowing _if_ the value of the current was positive. In other words the arrows define what positive and negative means.

For example, in many cases you don't know which way a current is flowing at the start. So you mark on the circuit an arrow pointing in any direction you like. Then you solve the equations and see what the sign is. Then you have to refer to the arrow to see what that means. In this case they are saying that _if_ I2 was a positive number it would be flowing up through R2. That doesn't mean I2 _is_ a positive number.

 

[phinds]

Doesn't Kirchhoff's Current Law state that the sum of currents at a node is 0?

It does indeed, which is why it makes no sense at all that you have summed 2 of the 3 currents entering the node and called THAT sum zero.

 

[rugerts]

Correct.
The arrows on a circuit diagram do NOT tell you the direction that current is flowing. They tell you which direction the named current (eg I1, I2, I3, I4...) would be flowing _if_ the value of the current was positive. In other words the arrows define what positive and negative means.

For example, in many cases you don't know which way a current is flowing at the start. So you mark on the circuit an arrow pointing in any direction you like. Then you solve the equations and see what the sign is. Then you have to refer to the arrow to see what that means.In this case they are saying that _if_ I2 was a positive number it would be flowing up through R2. That doesn't mean I2 _is_ a positive number.

Is this assignment of current direction similar to assigning directions to forces when using Newton's 2nd Law? That is, if you end up with a negative force, this just means that you've incorrectly assumed the direction of the force? Usually this meant redrawing the diagram to correctly position the force vector, does this also happen with circuits? Do I need to then redraw the current arrows?

 

[phinds]

Is this assignment of current direction similar to assigning directions to forces when using Newton's 2nd Law? That is, if you end up with a negative force, this just means that you've incorrectly assumed the direction of the force?

exactly.

Usually this meant redrawing the diagram to correctly position the force vector, does this also happen with circuits? Do I need to then redraw the current arrows?

No. Why bother? You have a current drawn and a value assigned and the two together show exactly what's happening.

You still haven't explained the zero.

 

[rugerts]

It does indeed, which is why it makes no sense at all that you have summed 2 of the 3 currents entering the node and called THAT sum zero.

Ok, I think I see now. So, currents 1 and 2 both sum to some new current, say current 3. I don't see a way for me to solve this with this new variable, current 3. Does this suggest that I should be instead using Ohm's Law instead of KCL to solve for current 2?

 

[phinds]

Ok, I think I see now. So, currents 1 and 2 both sum to some new current, say current 3. I don't see a way for me to solve this with this new variable, current 3. Does this suggest that I should be instead using Ohm's Law instead of KCL to solve for current 2?

No, it suggests you should do what one ALWAYS does with KCL in a 2-loop circuit. Write 2 equations in two unknowns and solve.

 

[CWatters]

Is this assignment of current direction similar to assigning directions to forces when using Newton's 2nd Law? That is, if you end up with a negative force, this just means that you've incorrectly assumed the direction of the force? Usually this meant redrawing the diagram to correctly position the force vector, does this also happen with circuits? Do I need to then redraw the current arrows?

You could but it's not necessary. It would be a mistake to do that in an exam as then your diagram and your working might appear or be inconsistent. If you were writing a technical manual for a piece of equipment it might make sense to redraw it to make it easier to understand and redo any calculations so its all consistent. A bit like avoiding "double negatives" when writing an essay.

 

[AVBs2Systems]

Hi.

There is just one voltage source, and there is a theorem that involves an ideal voltage source and all the elements to which it is parallel:
$$ \text{Theorem: An ideal voltage source in parallel to all other elements} \\ \text{determines the voltage drop across those elements, otherwise it would violate KVL and be outside the domain of simple circuit analysis} $$
You know the voltage in parallel is the same for those two branches of resistances, and you know the directions of the currents specified, so you can solve this, you have a current and a resistance given..
$$ \text{Corollary: You cannot have ideal voltage sources of differing magnitudes} \\ \text{in parallel, this would violate KVL.} \\ \text{Ideal voltage sources in parallel must be of the same magnitude and polarity} $$