# Homework Help: How do I find the current?

1. Jun 30, 2011

### joej24

1. The problem statement, all variables and given/known data

Number 8

2. Relevant equations

V = IR

3. The attempt at a solution

The points above the 3 horizontal lines represent grounded areas. They have zero electric potential there. How does the circuit work when it is grounded midway?

I found the total resistance to be 13 2/3
Since V = IR

24 = I * 13 2/3
I = 1.75

Why didn't this work?

#### Attached Files:

• ###### 001.jpg
File size:
23.6 KB
Views:
122
2. Jun 30, 2011

### Staff: Mentor

"Ground" is just a designation for a common reference node. Treat the grounds as though they are all connected by a single wire. That puts points A and B in the figure below at the same potential (they are in fact part of the same circuit node).

File size:
5.3 KB
Views:
205
3. Jul 1, 2011

### MalcolmKee

Since the two node is connected, short circuit does happen. Therefore your calculation is not sound.

4. Jul 1, 2011

### Quinzio

Short circuit does NOT happen.
A and B are actually the same electrical point.

5. Jul 1, 2011

### MalcolmKee

I know they are the same point.
That is the reason why electrical current will not flow to the upper side of the circuit. Instead, the current only exist in the lower mesh only.
Do I get it wrong?

6. Jul 1, 2011

### Quinzio

Nope, you're ok.
I mistook you as the OP.

7. Jul 1, 2011

### Ouabache

I suggest doing a KVL starting from the voltage source through the
resistor in question to ground potential. It may help if you redraw
your circuit in the standard fashion with voltage source at the left
and ground potential at the bottom.

8. Jul 1, 2011

### joej24

What's a KVL?

So the grounds can be combined to be one single point? Can you explain this to me?

How do we know if it short circuits or not either?

This is my attempt at redrawing the circuit.

File size:
2.3 KB
Views:
79
9. Jul 1, 2011

### joej24

How am I supposed to start this problem? My physics class didn't cover any E and M material. Bear with me because I really don't know how to do this.

The grounding part made the entire problem confusing to me.

10. Jul 1, 2011

### Ouabache

Sorry, if they have not covered Kirchhoff's Voltage Law in your course yet, it may be useful to look ahead to see what it means.
It makes solving this question, (as one of my former profs used to say a lot) trivial.

Basic Definition: The sum of the electrical potential differences (voltages) around any closed circuit is zero.
This implies that the sum of the applied voltages in any closed loop is equivalent to the sum of the potential drops in that loop.
Hint: the 2nd ground symbol is at the same potential (0v), as the one at the negative pole of the voltage source.

Here is one http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Basic/Basic5Kv.html#Writing" with good examples of how to use it, under "Writing the KVL Equations".
Here is a recent discussion by a PF member of https://www.physicsforums.com/attachment.php?attachmentid=36230&d=1307457441". (it was attached as a pdf file).

With a basic understanding of KVL, you can solve this question using Ohm's law, a law you already know.

Not quite, I suggested putting the voltage source at the very left side. As you've redrawn the circuit, you've left out the ground connection between the 8 ohm resistor and the rest of the circuit. Your combined resistances are correct.

Last edited by a moderator: Apr 26, 2017
11. Jul 2, 2011

### joej24

So do Independent Loops start with the battery or any source and end at the battery/grounded point? In upper div. physics are there multiple batteries to make circuits more complicating?

Grounded points have 0 V too right?

I see two loops here.
Loop 1: Battery, R1, R2, R4, R5, back to bat.
Loop 2: Battery, R1, R3, R4, R5, back to bat.

I guess we could use kirchev's law.
Loop 1: VB - I1*R1 - I2*R2 - I4*R4 - I5*R5 = 0

Loop2: VB - I1*R1 - I3*R3 - I4*R4 - I5*R5 = 0

But if Grounded points have 0 V, From the battery to the G1 = 0.

VB - I1*R1 = 0
24 - I*8 = 0
I = 3 A.

And since they are asking for the current through R1, 3 is the answer no?

And just to make sure, I1 = I4 = I5 right?

Last edited: Jul 2, 2011
12. Jul 2, 2011

### joej24

Opps. I forget to attach the REVISED circuit drawing :D

I think there should be currents labeled between R 1 and R 2/ R 3 and between R 5 and the battery, but they aren't really relevant.

#### Attached Files:

• ###### circuit rev.png
File size:
14.5 KB
Views:
94
13. Jul 2, 2011

### Staff: Mentor

Since points A and B are one and the same node (they are connected via their "ground" connections), no current at all will flow through any of the resistors above the A,B points in the diagram I modified, because there can be no potential difference between points A and B. Therefore, concentrate your efforts on the 24V battery and the 8Ω resistor, which form a series connected loop.

14. Jul 2, 2011

### Ouabache

Voltage source (e.g. battery) is typically (but not always) attached to ground potential at its negative terminal. Circuit loops can but do not have to include voltage sources.

yes, there may be multiple batteries in a circuit, which may look more complicated, but when you become used to applying KVL, you will see, they are not that hard to analyze.
Yes, the grounded points are at 0 voltage potential.

That's right

If G1 was not there, these loop equations would be valid.
If you follow the discussion gneill described,
you will note that the short circuit simplified your problem.

The pdf file I pointed to in post #10 was contributed by stevenb

Last edited: Jul 2, 2011
15. Jul 3, 2011

### joej24

How does the circuit short circuit? Since the electric potential is 0 at the grounded point doesn't that just mean that there is no energy to push up the charge through the upper part of circuit? So rather, charges go to the ground and back to the battery.

The definition of short circuit by wiki

"short circuit is an abnormal low-resistance connection between two nodes of an electrical circuit that are meant to be at different voltages. This results in an excessive electric current (overcurrent)"

16. Jul 3, 2011

### Staff: Mentor

Potential energy is provided by the battery. Ground is just a reference point, and does not actively push, pull, or dissipate current in an fashion. You could connect points A and B with a wire (as I did) and erase the ground symbols entirely, and the circuit behavior would be the same.

The wiki definition of short circuit that you quoted is limited and a bit naive. There is no reason why an "excessive" current must flow. A short circuit in a low voltage, low current environment is still a short circuit. Also, there need be nothing "abnormal" about the resistance comprising the short; sometimes short circuits are deliberate (like when you're solving for Thevenin or Norton equivalents). Granted, there are situations where a short circuit is bad news -- such as when dealing with car batteries or household wiring where the shorts are usually unintentional and can have undesirable outcomes-- but shorts are not always a bad thing.

Originally, a "short circuit" was "short" because it provided a shorter path for current to flow, depriving a longer path through the network of both voltage and current. For purposes of discussing circuit analysis, just think of a short circuit as tying together two places in a circuit that would, without the "short", be separate nodes.

17. Jul 3, 2011

### Ouabache

here I redraw your original circuit (see attachmt)
I mentioned it is useful to redraw your circuit by standard convention
(voltage source at the left, ground at the bottom) as you can often
analyze them more easily.
Analogous to gneill's drawing, I have connected the two ground points you labelled G1 and G2
(between nodes A and B). You can easily see there is a shorter path for current to flow to ground through the single line A-B, versus going through the rest of your circuit to the right of A-B.
Since the resistance of that wire is very small, virtually all of the current will flow along that path.
An so the current through all the circuit to the right of A-B is virtually 0 A.

That is why you can exclude the impedance network on the right side of A-B, from your analysis, and as I mentioned earlier (post #10) , you're left with a simple circuit that can be solved using Ohm's Law, I = V/R.
With a basis knowledge of Kirchoff's Laws, you found the current even more easily
as you described above.

In your circuit you don't have "excessive current" through the short because it is limited by the 8 ohm impedance that precedes it. You only gave part of the wiki's definition of short circuit. You left out an important part.

#### Attached Files:

• ###### ouab_01.jpg
File size:
16.8 KB
Views:
87
Last edited: Jul 3, 2011
18. Jul 3, 2011

### joej24

Okay. Thanks everyone. I understand the problem now. I have just one last question. Why does the current NOT flow through the wire past the grounded point?
When grounding occurs, does all the current/charges flow to the ground, leaving none for the part of the circuit past the grounded point?

I know the electric potential turns to 0 at grounded points. The ground is connected to the negative terminal of the battery, so the loop continues. There is no potential to do work and push charges past the part of the circuit past the grounded point.

Is this correct?

Last edited: Jul 3, 2011