Ok i'm not sure if I did these right, and my teacher isn't exactly sure how it's done either. Here are the problems: 1. Find the current that flows through the [tex]3\Omega[/tex] resistor. 2. What is the potential of point D relative to point C? This is what I did for #1: I solved the parallel branch into a series equivilent by finding the resistance equivilent of the 3, 4, and 5 [tex]\Omega[/tex] resistors and got [tex](\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{-1}=1.27\Omega[/tex]. Since all the resistors were in a series now I took the sum and got [tex]R_{eq}=1.27+2+ 1=4.27\Omega[/tex]. To find the current throughout I did [tex]\frac{V}{R_{eq}}=\frac{12}{4.27}=2.8A[/tex]. Now I had to find the current in the [tex]3\Omega[/tex] resistor so I think I had to find the voltage at the point before it went into the parallel circuit which came out to be [tex]V_{3\Omega}=\xi-(V_{d2\Omega}+V_{d1\Omega})=12-((2.8*2)+(2.8*1))=3.6V[/tex] and then divide that by 3 to get 1.2A For #2: I had the same problem on this one as the last one. Basically on how to find the current in a parallel circuit. I know Kirchhoff's junction rule in that the sum of all the currents going in, must be the sum of all the currents coming out, and I know that the voltage stays the same and current changes in parallel circuits. I've heard that the current is split evenly between each branch, is proportional to each branches resistance, and/or it is calculated by the voltage divided by the resistance at that point. I'm not sure which, if any, to use. My answer, which stays consistant with #1 of 8V is probably not correct since that would mean more voltage than the terminal. But if the split between the two branches it stays consistant with Kirchhoff's rules and not #1 My teacher didn't go over anything about RC circuits, just that the resistance is calculated in the direct opposite manner of resistors. I think RC circuits might be on the AP exam, and my book doesn't have any RC circuit problems. Anyone know where I can find some? Also does anyone know what AP Physics B is in college?
Your work on the first part is correct. If you repeat the methodology on the 2nd you should get the correct result. As you are aware your answer for the second part is not correct. In the second problem since the resistance of each branch is the same, 6 ohms, the current through each branch will be the same. Remember the voltage drop across a single resistor is determined by the resistance and the current.
Your solution for the first problrem is correct. Just apply the same method for the second one. Both 4 ohm resistors are connected in series, the voltage accross them is known, so you get the current. Multiply it by 4 ohm, you get the voltage across the resistor CD (3 V) The voltage is obtained as potential difference. VDC = UD-UC. You need the potential of D (UD) with respect to UC. The magnitude is 3 V, you have to decide the sign. A is connected to the positive terminal of the battery, so A is more positive than C. The current flows in the direction A->D->C. The current accross a resistor flows along decreasing potential, so D is more positive than D, that is the potential of D with respect to C is 3 V. ehild Sorry, I misread the 2 ohm resistor... so both the voltage and potential are 4 V.