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Circuits and Resistance 
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#19
Sep2613, 10:49 PM

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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) 


#20
Sep2613, 10:50 PM

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



#21
Sep2613, 10:51 PM

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#22
Sep2613, 10:55 PM

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#23
Sep2613, 11:00 PM

P: 178




#24
Sep2613, 11:02 PM

P: 178

* i took different values for the resistors and voltage , i think it depends on the values then * 


#25
Sep2613, 11:03 PM

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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. 


#26
Sep2613, 11:07 PM

P: 178




#28
Sep2613, 11:11 PM

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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. 


#29
Sep2613, 11:18 PM

P: 178

but i apologize i made a stupid mistake , i am sorry again 


#30
Sep2613, 11:19 PM

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#31
Sep2713, 07:31 AM

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. 


#32
Sep2713, 07:50 AM

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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. Since the possible answers are: brighter, dimmer, stays the same ... there is a 1/3 chance of getting the right answer just by guessing. It follows that the reasoning that is followed needs to be clear and correct to get good marks. 


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