Ive been struggling with Physics. Can someone help explain this electrical current and circuit problem?

Question

Figured id take a shot. If anyone can help with the physics problem it would be much appreciated!Thanks

Two circuits (the figure below) are constructed using identical, ideal batteries (emf = ) and identical lightbulbs (resistance = R). If each bulb in circuit 1 dissipates 5.0 W of power, how much power does each bulb in circuit 2 dissipate? Ignore changes in the resistance of the bulbs due to temperature changes.

-----------------
l---------------l
l---------------l
-----(+)--------l
----(-)---------R
l--------------l <----- Circuit 1
l--------------l (Resistors are in series)
l--------------R
l --------------l
__________________
l---------------l---------------l
l-----(+)-------l--------------l <---- circuit 2
l---(-)----------l---------------l (Resistors are in parallel)
l------ Resistor---- Resistor
l---------------l---------------l
l---------------l---------------l

Answer 1

Hard to tell what the circuit is. Let's pretend the first circuit is n bulbs with resistance R each and a battery with voltage V all in series.

The effective resistance Re = R+R+R+..+R = nR. So the current is :

I = V/(nR) now the power each bulb dissapates is P = I^2R

So P = V^2/(n^2R) or V = n sqrt(RP)

Now lets suppose the second circuit has the n bulbs all in parallel to each other and the battery: The effective resistance is:

1/Rp =1/R +1/R +...+1/R = n/R --->Rp =R/n

Now each bulb has the same voltage V across it so the power the bulb dissapates is:

P1 = V^2/R = n^2 R P/R = n^2 P

You were given P and presumable N so you can find P1

Answer 2

Calculate the clever resistence over the resistors in parallel: a million/Rt = a million/R1 + a million/R2 = a million/12 + a million/4 = 4/12 = a million/3 So the voltage over the parallel resistors is the equivalent of the voltage over a 3ohm resistor. V = IR, so I = V/R = 20/(3+7) = 2. So i could say 0.5 A is going by the 12 ohm as that's thrice larger than the 4ohm resistor.