# Potential to earth with 3 power sources

Somebody who can help me calculate the potential to ground in point a,b,c and d? I would appreciate a good explanation, because my head gets all messy when the circuit has more than one power source.

Have worked out: A=6V, b=14V, d=0V, but I'cant figure out C. I tried to calculate the voltage, which I set to 14v-8v+6V = 12V. Then I found the current in the circuit to be 0,6 A, and the voltage drop across each resistor to be 6 V.

• This looks like a homework question. What have you tried so far? Can you work out points b and d? They are the easiest to do first. Then try a. Oct 11 '16 at 12:35
• You'll need to show some effort for a homework question. How can anyone know where you're getting stuck otherwise? Just giving you the answer achieves little. Oct 11 '16 at 12:38
• Have worked out: A=6V, b=14V, d=0V, but I'cant figure out C. I tried to calculate the voltage, which I set to 14v-8v+6V = 12V. Then I found the current in the circuit to be 0,6 A, and the voltage drop across each resistor to be 6 V. The problem is that I get different a potential in C if I go one way to ground, or another. So thats my problem, and believe me, I have done a lot of research to find out! Sorry for not posting my thoughts about the circuit from the start
– JRB
Oct 11 '16 at 12:45
• Now, the voltage you've calculated is not right, that's why it's not working. Can you work out the voltage at unlabelled point above the 6V cell and below the vertical resistor? That's the next step. Oct 11 '16 at 12:54

For C you have 14V and 6V in series making 20V. So the current through the resistors is 1A. making a voltage drop of 10 V over each resistor. So the voltage at c would be 14 -10 = 4V. Or along the other path 10 -6 = 4V also. Look at the polarity of the sources so you can correct the error. Thereafter it is easy

The 8 V source is not inside any circuit so no current can flow.

• Thanks. But how can u totaly look away from the 8V source, wouldn't it effect the voltage in point b?
– JRB
Oct 11 '16 at 13:18
• No. No see my adjusted answer. Oct 11 '16 at 13:20
• @JRB (b) is ideal Voltage source. Nothing can alter (b) `"with respect to" (w.r.t.) ground. Only loop KVL rules apply Oct 11 '16 at 13:35

Now if you want to find c, you can omit the 8 volt battery because it does not affect the part of the circuit you are looking for. That ground symbol references 0 and you found out that b is 14. Now the point between the 6 volt battery and the 10 ohm resistor is -6 volts since the battery is the wrong way around. Now you got a voltage divider between the resistors so c must be halfway between 14 and -6 volts.

Basically simplified and easier to understand.

All that matters for C is that there is a closed circuit. So Kurchoves Voltage law means that going around in a circle from any point back to the same point, MUST result in a voltage of 0.

Also V=IR.

So starting at d (because it is common zero volts) and going clockwise: 14V + (10*I) + (10*I) + 6V = 0 (where I is current through resistor).

0 = 20V + 20*I.

I = -1;

V=IR. So each resistor has -10V across it ( you drop 10V moving from one side of the resistor to the other side when you move clockwise through the circuit).

So if you move from common to point C, in clockwise direction, you go from 0V up to 14V then cross the resistor is -10V, arriving at 4V.

All of this can also be confirmed by going "anti close wise" around the circuit:

-6V + 10*I + 10*I - 14V = 0

So I = 1, So voltage across resistor is +10V in anti clockwise direction (you gain 10V when moving from one side of the resistor to the other side in anti clockwise direction).

So moving from common to C: 0 - 6V + 10V = 4V.

Both ways tell us C is 4V with respect to common 0V.