i got a Req = cde / (de + ce + cd) just by combining the cde resistors.. im still not seeing it

When you close the switch you're adding another path of current. Voltage does not change because the resistors are in parallel, and I must increase because Itotal = Ic+Id+Ie, therefore resistance must decrease (ohm's law V=IR)

Also the equivalent resistance is always less than any of the individual resistors in parallel

This post was edited by drxscillator on Apr 19 2013 07:22pm

r eqv = r + 1/(1/r + 1/r +1/r + .... )

add more-> smaller r eqv

so by adding E (another resistor) current is drawn more to light up the lightbulb therefore making overall resistance decrease?

so wat i said was correct than? the brightness of A will decrease because more current is drawn in order to power up E.

but how do u know all the lightbulbs have same resistors?

This post was edited by FamilyGuyViewer on Apr 19 2013 07:47pm

I was just saying resistance = R for theoretical purposes. Anyways, the brightness of A will increase because the current of the circuit increases because the equivalent resistance (of the whole circuit) has decreased when a third resistor was added in parallel. (Power absorbed by a resistor = I²*R) Resistance and current have an inverse relationship in parallel

This post was edited by drxscillator on Apr 19 2013 08:09pm

all the lightbulbs have the same R?

no I was just using that to prove that adding resistors in parallel decrease equivalent resistance

but than that only works if all the lightbulbs are the same R though... it doesnt say all the lightbulbs have same R, so that analogy wouldnt work for this problem

That principle applies for any resistance of R, although you're right that analogy was not that great. Think of closing the switch as adding another path for current to flow, when you add another path for current to flow the total current produced by the emf increases and due to ohm's law when current increases resistance must decrease in order to maintain a constant voltage.