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Hello! In the figure above I was asked to determine the potential difference between the points x and y ( I had this in my exam yesterday). Here what I did: The potential difference from x to A is: V= IR = 2*4 = 8v The potential difference from B to y is: V= Vb+IR =14+ (4 *1) = 18 OK, Then I should determine the potential difference from A to B, and here I got confused, I didn't know what to do, I supposed that the the P.D. on this area would be 12V because (3*4 =12 ) but this is on the wire which doesn't have a battery on it, but what about the other wire?! Could you please help me understand how I could solve it and the explanation for that? Thanks in advance! Note: In my studies I consider that the current flows from the positive pole to the negative.

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You need to consider only one path from x to y to find the potential difference. So you sum the potential differences along the path via the 4 Ohm resistor.

So from left to right you get: +8V + 12V + 14V + 4V = 36V. The last voltage drop over the 1 Ohm resistor is 4A*(1 Ohm) = 4V. The current flowing out of the circuit at y through the 1 Ohm resistor has to be equal to the current flowing in at x through the 2 Ohm resistor. The other path gives the same potential difference when you take the 1V potential drop over the 1 Ohm resistor and add a battery voltage of 11V.

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  • $\begingroup$ OK, but if the battery of 14V was connected with the battery of (Vb)2 on parallel, I mean we will put it in an opposite direction in the circuit , would that change the potential difference? $\endgroup$ – Asmaa Sep 13 '16 at 11:59
  • $\begingroup$ There is no change because the currents and voltages are given in the path via the resistor only. $\endgroup$ – freecharly Sep 13 '16 at 16:31

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