# Electric current and direct current circuits

1. Sep 17, 2006

### Trifecta

I seem to be having a problem getting this. I can get the first question, it is .65 A. But I can't seem to figure out the potential at point B. I used Vab = I2 * I1 but it still doesn't work.

2. Sep 18, 2006

### andrevdh

Point B is effectively connected to the positive terminal of the battery. So I would think that is would be at +15 V with respect to ground.

3. Sep 18, 2006

### Integral

Staff Emeritus
The trouble here is that ground is NOT the negative terminal. You need to reduce the series-parallel resistor network to its equivalent between A and B. You can then compute the total resistance thereby finding the total current. Once you have that you can easily compute the voltage drop across each resistor.

4. Sep 18, 2006

### andrevdh

Would $V_B$ then be $V_{AB}$, that is the sum of the potential drops over the 6 and 12 ohm resistors? Which should then be the same as that over the 7.5 ohm resistor? Which means that the ground just defines a zero reference level?

5. Sep 19, 2006

### andrevdh

I investigate it experiementally and it seems that connecting the circuit to ground has no effect on the current in the circuit of the potential difference over the components. This lead me to think that answer to b should be the positive potential over the 7.5 ohm resistor. It seems grounding part of a circuit just defines a zero reference. It is very much like ground acts as a floating potential. It can assume any potential. Any electronic boffins out there?

6. Sep 19, 2006

### FunkyDwarf

i always thought you couldnt get voltage at a point, because its a difference not a specific value

7. Sep 19, 2006

### andrevdh

Good point (pun not intended - I always wanted to say that). But what when I tried to measure the potential between the terminals of the power supply and earth the meter indicated no potential difference!

8. Sep 19, 2006

You can get a voltage at a point, but the point has to be relative to something. Often a node is assigned, call it $V_g$. This value is set to 0. Thus, when a voltage is specified it is given with respect to this ground.

I think the easiest way to "see" this circuit is as follows:

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9. Sep 19, 2006

### FunkyDwarf

ok, point taken. :)

10. Sep 20, 2006

### andrevdh

It seems that there is a difference between earth and ground as far as electronics is concerned. In electronic circuits it is a common practice to connect components to a common ground, which is the negative terminal of the power supply.

This means that point B get connected to the negative terminal of the battery and it becomes the reference for the potential at B. That means that the 11 ohm resistor is eliminated and the required potential is that over the parallel circuit formed by the I1 and I2 branches.

Last edited: Sep 20, 2006
11. Sep 20, 2006

### semc

I tried out the question and found out that the potential at B is 4.88 ohm is that right?

12. Sep 20, 2006

Almost.

The units are Volts :)

...good otherwise though

13. Sep 20, 2006

### andrevdh

With point A connected to ground we have the total potential of the battery applied over the remaining circuit. That is both points A and B connected to the battery so that the potential over B will be that of the battery.

Last edited: Sep 21, 2006
14. Sep 21, 2006

### semc

Can you just find the effective resistance across all the 3 resistor on the right and use potentiodivider to find potential drop across the effective resistor. After finding the P.D across, use V=IR on the 7.5 ohm resistor to find P.D across?

btw, i was trying to say 4.88V sry bout that

15. Sep 21, 2006

### andrevdh

I think it is much easier than that. Notice that point B is connected to the + terminal of the battery. When A is connected to ground it gets connected to the - terminal of the battery. So the total potential of the battery is applied to the circuit. This means that B will be at +15 V with respect to ground (the negative terminal of the battery).