When the switch is closed, what happens to the brightness of bulb D? What happens to the brightness of bulb A? Explain your reasoning. I initially thought it would decrease because current splits between D and BC. But since it is parallel to BC would it increase because the it would be R/2 instead of R? And will A stay the same?
Let's assign some resistance values and voltage. Let's say that every bulb has a resistance of 10 ohms and the voltage is 10 volts. Now, find the current and voltage drop across each bulb before and after the switch is closed. Do you know how to do that?
notice that when you connect a resistor in parallel , it gets a lower value than the value of the resistance itself that means if you have one 2 ohm resistor , there you go you have 2 ohms of resistance , but if you connect 2 ohms with another 2 ohms in parallel , you get a 1 ohm equivalent resistance , when you close the switch , D is now parallel connected with B and C , thus it has lower resistance everything lights up because now you have alot less equivalent resistance , so more current is present but also current is split up between D and b - c i think that the current split over comes the decrease in resistance thus decreasing the its intensity , but E surely gets brighter , A surely gets brighter , D gets dimmer if they all have the same resistance
Hmmm... sort of, but how? If I double the resistance but halve the current will that make the bub brighter or dimmer? Is there a single quantity that the bulb brightness is related to? In this case you can probably get away with modelling all the resistances as a single value ... so if I double the current and keep the resistance the same, what happens to the brightness? Do you know how to calculate the current through the bulb in each case? Like Drakkith is pointing out, it is possible to do this by rules of thumb: looking at the possible paths for the current in both cases. This requires you to understand how electric circuits work. When you don't have this understanding, you have to do it the hard way :( Note: I'm just going to let him talk to B4ssHunter. "Current Intensity" is the old name for "Electric Current" which we just shorten to "Current".
Thinking the basic way i just though a would be the same and d would be dimmer. but thinking quantitatively i got confused
Assuming values are set per your post #5, shouldn't that be 0.5 amp for A and D, and 1 amp for E? A and D in series is 20 ohms.
Ah! That's what I get for second guessing myself! Plus the 2 hours of math homework I've already done.
How do you get that? If you show me your calculation, I may be able to help with that. It would also help if you answered the questions put to you in the thread. I have a feeling you are not the only one getting confused ;)
So since A= BCD. it wil be brighter. and since the current is split between that then half goes to D then half goes to BC. so thats how you get it. But E would stay the same I believe because it is its own independent pathway....?
E is in parallel with everything else and has nothing in series. Therefor the voltage across it is always 10 volts. As such, closing the switch will do nothing to it. The current will remain at 1 amp. A and D are in series, with 10 ohms of resistance. This adds to 20 ohms for a current of 0.5 amps through their leg of the circuit. Now, when the switch closes, we have 2 more bulbs added that are in series with each other, parallel to D, and in series with A. The resistance of leg BC is 20 ohms. Now we find the equivalent resistance for BCD, and after knowing that we can figure out the voltage drop across A and BCD and the current flow through that leg (all of which passes through A). THEN knowing the voltage drop across BCD, we can find out how much current is going through BC and D. At that point you will have the current through each bulb. (It's been a while since I took my basic electronics course, so someone correct me if I'm wrong)
according to my calculations , when the switch is closed , total resistance decrease thus current increases , but also the resistance of the first branch drastically decrease so it takes more current so it dims E , only A would be brighter D dims and ofcourse the other two bulbs would just light up