# Circuits and Resistance

by disneychannel
Tags: circuits, currents, resistance
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P: 11,618
 Quote by disneychannel BCDE would have 1/3 resistance so 3 times the current
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)
 P: 178 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
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P: 11,618
 Quote by disneychannel 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....?
BC is twice the resistance of D and will not get half the current through that leg.
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P: 11,618
 Quote by B4ssHunter according to my calculations , when the switch is closed , total resistance decrease thus current increases
TOTAL current is irrelevant for this problem.

 but also the resistance of the first branch drastically decrease that takes more current such that it dims E , only A would be brighter D dims and ofcourse the other two bulbs would just light up
No, the voltage applied to E remains at 10 volts, just as the voltage across leg ABCD is 10 volts. Since you have 10 volts applied to 10 ohms, you have 1 amp of current across E at all times.
P: 178
 Quote by Drakkith TOTAL current is irrelevant for this problem. No, the voltage applied to E remains at 10 volts, just as the voltage across leg ABCD is 10 volts. Since you have 10 volts applied to 10 ohms, you have 1 amp of current across E at all times.
i used another set of calculations not the ones you are using
P: 178
 Quote by Drakkith TOTAL current is irrelevant for this problem. No, the voltage applied to E remains at 10 volts, just as the voltage across leg ABCD is 10 volts. Since you have 10 volts applied to 10 ohms, you have 1 amp of current across E at all times.
yes the voltage remains the same , but the branch now has less resistance , so more current goes through the branch , even though since the total resistance decreased more current goes through E , but it does not compensate the loss of current in E resistor due to the decrease in resistance in the branch
* i took different values for the resistors and voltage , i think it depends on the values then *
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P: 12,482
 Quote by disneychannel 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....?
That's right, E stays the same.

But I don't know what you mean by "A=BCD".
However: the current only splits in half if there is equal resistance on both sides.

P: 178
 Quote by Simon Bridge That's right, E stays the same. But I don't know what you mean by "A=BCD". However: the current only splits in half if there is equal resistance on both sides. Note: if you don't answer questions I cannot help you.
how does E stay the same ? when the parallel resistors were connected , the whole resistance of the branch was decreased , thus more current would flow through the branch and less current would flow through E , even if the intensity went a little bit higher due to the total decrease in resistance
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P: 11,618
 Quote by B4ssHunter yes the voltage remains the same , but the branch now has less resistance , so more current goes through the branch , even though since the total resistance decreased more current goes through E , but it does not compensate the loss of current in E resistor due to the decrease in resistance in the branch
E's resistance is NOT changing.
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P: 11,618
 Quote by B4ssHunter how does E stay the same ? when the parallel resistors were connected , the whole resistance of the branch was decrease , thus more current would flow through the branch and less current would flow through E , even if the intensity went a little bit higher due to the total decrease in resistance
You're coming at this from the wrong angle. Do not look at resistance changing. The resistance of any component will NOT change.

As such, if the resistance of E doesn't change, and the voltage drop (which is equal to the applied voltage) doesn't change, then current flow MUST stay the same through E.
P: 178
 Quote by Drakkith E's resistance is NOT changing.
i know it doesn't
but i apologize i made a stupid mistake , i am sorry again
P: 178
 Quote by Drakkith You're coming at this from the wrong angle. Do not look at resistance changing. The resistance of any component will NOT change. As such, if the resistance of E doesn't change, and the voltage drop (which is equal to the applied voltage) doesn't change, then current flow MUST stay the same through E.
btw when i said the branch i meant the one without the E , the one containing a , b , c and D . not E i know E is not changing
 P: 109 I assigned the voltage to V and the resistance of each light to R. Using V = I * R, parallel resistance equivalence, and current division I was able to algebraically determine current through the ABCD leg and then the voltage drop through A and D. Skip intuition on this one it trips you up.
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P: 12,482
 Quote by 2milehi Skip intuition on this one it trips you up.
I noticed that too - both the current and the voltage changes.

What I have been trying to get OP to realize is that it is the power dissipated in the bulb that determines the brightness. This is proportional to the square of the current or the inverse square of the voltage.