Odelouche 16 Report post Posted April 27, 2016 Hi Guys, I'm having a hard time understanding this question: I failed it miserably in my mock, then went back, studied the concepts, tried again, and again couldn't come up with the right answers. I don't really understand what the mark scheme is doing, an explanation of the equations used as well as how the direction of current is determined and used would be much appreciated. The question and mark scheme are attached, thanks a lot for your replies! Share this post Link to post Share on other sites

kevG 160 Report post Posted April 27, 2016 Hey, I'll try my best to explain it to you and you'll realize that it is very simple. In the end it is only a play on numbers through simultaneous equations. We know from the Kirchoff's first rule is that the sum of currents going through one junction is equal to the sum of currents flowing out of the same junction. You could do this anyway you want and I'm pretty sure you'll come up with the same answer. This means that at the X junction, when a current i1 is going inside the junction, the corresponding currents coming out should be equal to i1, which gives us i1 = i2 + i3. This is the application of the first rule which we can use later on. After getting that equation we can move onto Kirchoff's second rule which tells us that the direct summation of all voltages in a loop should be equal to 0. We know that in the first loop, the cell has an emf of 9V and it is only possible for that loop to have a total voltage of 9V and this is split among the resistors in that loop which would give something like 9V = 3i1 + 6i3 (we know the voltage using V = IR) We have to do the same with the second loop with the cell emf of 12V, but remember that current flows from the positive to negative terminal, the way I defined i2 in the second loop is relative to the first junction, so i2 should be negative when we're talking about the second loop as conventional current moves from positive to negative. We then end up with something like 12V = -2i2 + 6i3. After finding the 2 voltage equations, we only need to solve for i3 using simultaneous equations and then boom, we get the current across the resistor with 6 ohms. I attached my working with a little bit of explanations here and there to help you out.. Oh and you will notice that I get a positive value for i3, but this is only because I used the X junction instead of the Y one as shown in the markscheme, but in the end they both make sense as a positive answer tells us that the current moves from X to Y, which is exactly what the markscheme tells us. Hope I helped and ignore the grammatical or spelling mistakes... It's like 2am here and I'm a bit slow Peace 2 Share this post Link to post Share on other sites

Guest iblearner Report post Posted May 1, 2016 Hope this helps! Coincidentally, I did this question just last night! Share this post Link to post Share on other sites