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.
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.andrevdh said: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.
You can get a voltage at a point, but the point has to be relative to something. Often a node is assigned, call it [itex] V_g [/itex]. This value is set to 0. Thus, when a voltage is specified it is given with respect to this ground.FunkyDwarf said:i always thought you couldnt get voltage at a point, because its a difference not a specific value
Almost.semc said:I tried out the question and found out that the potential at B is 4.88 ohm is that right?