Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys
1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.
2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.
For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.
You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.