Kirchhoff's law question?

Summary if you already understand KVL:

Basically just start by assuming a polarity and then if it's wrong, you'll get a negative number, and then you just flip the polarity.

Long version which explains KVL:

Take any point on a circuit. That point has a certain potential with respect to the ground. When you go along the circuit, you will experience different rises and falls in potential as you cross different circuit components, but if we make a full cycle then we have to get back to the same potential at our given point. This means that the net potential gain around any loop has to be zero. This is Kirchhoff's Voltage Law, abbreviated as KVL and also known as the loop rule.

So, imagine starting at point a and going counterclockwise. You will gain 14 V going across the first battery, then you will gain 6 V across the next voltage source, then you will lose 2 V going across the next source (mind the polarity) and finally, if we initially assume the unknown source is oriented the same way as the 14 V battery, we gain Vx. We can safely assume a polarity here because if we guess wrong, we'll just get a negative voltage, which indicates that we guessed wrong and the polarity goes the other way.

So we have that 14 V + 6 V - 2 V + Vx = 0, so Vx = -18 V. Again, don't worry that we get a negative answer, that just means we need to flip the polarity and the potential difference is 18 V with the positive terminal on the right.

Hope this helps, and happy new year!

where were you guys when i was in the college?

i never understood then,

now things couldn't have been made clearer or simpler.

thanks every buddy.

Va = -20 volts, which is the sum of -6 an -14 volts

Vb = - 2 volts

So Vab = Va - Vb = -20 -(-2) = -18 Volts

6v+14v=-20v @ a

-20v-Vx+2v=0>>-20+2=Vx

Vx=-18v

At ground voltage=0. 14+6 is the voltage= voltage ab +2 This gives voltage ab= +18 V

sum of the voltages around a loop has to be zero.

taking CCW as +, starting at battery, and taking Vx positive as point a, we have

+14+6–2+Vx = 0

Vx = –14–6+2 = –18 volts

or Vab = –18 volts

or Vba = +18 volts