for the last question, i was thinking combining the series resistor (4 ohm and 8 ohm resistors) so than i would have a parrallel circuit of 12 ohms, 6 ohms and 3 ohms respectively.
than combine the parrallel circuits (12ohm, and 6 ohm) which would lead to an equivalent resistor of 4 ohms.

it would look like a batter ont he far left side, than a 4 ohm resistor down the middle and the 3 ohm resistor and the capacitor on the right side.

can i do tat?



for first question, the order would be A > B > C = D. all the current will hit A, than the current will split into B and and split again between C and D.
adding the E lightbulb wouldnt affect A because its in parallel circuits?

This post was edited by FamilyGuyViewer on Apr 17 2013 03:13pm

Can't combine the 12/6/3 ohm cause they are NOT in parallel because of voltage drop from EMF/capacitor

probably just use kirchoff's loop rule to solve the entire circuit since you have all the necessary information - then find what you need from there

you're right for 1st one

This post was edited by silvermace on Apr 17 2013 10:21pm

my entire answer for Q1 is right?

up

Why do you think adding another won't affect A?

Ask yourself what does it do to the overall resistance , i=v/r
v=const

power=vi

Adding E in parallel would decrease the eq. resistance

the re resistance of which lightbulbs?

hmm maybe adding E will decrease the birghtness of bulb A but it would have the same brightness as C and D because of pararel circuits. if this is the right explanation then idk how to put it in terms of the equations

I'm talking about the equivalent resistance of the circuit in general. If R,eq goes down then the total circuit current increases (I = V/R, assuming the emf is constant). Now that the current is increased it will make A brighter(P = I^2*R). Then the current will split between B and the CDE(which will split again across each light bulb) parallel connection. Since the total current has increased the current across B will increase making it brighter, but because the current is being split to CDE instead of just CD the 3 in parallel will become dimmer.

I'm not sure how understandable I've made this but this sounded much better on paper

as far as the equation
assuming each resistor is equal to R
Switch open
C and D in parallel
1/Rcd = 1/RC+1/RD = 2/R ---> Rcd = R/2
B and CD in parallel
1/Rp = 1/RB + 2/R [1/Rcd]= 3/R --> Rp = R/3

A and Parallel bulbs in Series

R + R/3 = 4R/3

Switch Closed
C D and E in parallel
1/Rcde = 1/RC + 1/RD + 1/RE = 3/R --> Rcde = R/3
B and CDE in parallel
1/Rp = 1/Rb + 3/R [aka 1/Rcde] = 4/R ---> Rp = R/4

A and parallel parts
R + R/4 = 5R/4


This post was edited by drxscillator on Apr 19 2013 04:22pm

umm how do u know Req of the circuit in general will decrease if E is added?

Reciprocal formula, the more resistors you add in parallel the lower the equivalent resistance.