# Homework Help: Circuit brightness problem

1. Aug 18, 2007

### ehrenfest

Can someone help me with 1 (e) at the following site:

http://ocw.mit.edu/NR/rdonlyres/Physics/8-02Electricity-and-MagnetismSpring2002/29B6C858-A92B-431F-A43E-725D58A768EF/0/exam1.pdf [Broken]

The solutions say that all on-bulbs in 1 (e) glow with equal brightness. But how is that possible because in e-c, the equivalence resistance is

(2/R) so the total current is V * R/2 and the current through each branch is V * R. In e-b, the current is V/R through B, which is different, right?

Last edited by a moderator: May 3, 2017
2. Aug 18, 2007

### learningphysics

2/R as the equivalent resistance is wrong. It should be R/2. You don't need to find the equivalent resistance though. The current through a resistor is just the voltage across the resistor divided by resistance. For e-c, the voltage across bulb B is V... the resistance of bulb B is R, so the current through bulb B is V/R... exactly the same way, the current through bulb C is V/R.

Last edited by a moderator: May 3, 2017
3. Aug 18, 2007

### ehrenfest

I see. But you can just keep adding more and more light bulbs in parallel and they will each glow as bright as a single light bulb? Doesn't that allow you to generate an infinite amount of light (energy) with a finite amount of voltage?

Yes, we are assuming the wires have negligable resistance. But this still seems fishy to me...

4. Aug 18, 2007

### mgb_phys

They are in parallel so it's an infinite amount of light with an infinite amount of current. This isn't a problem as an infinite amount of current at a finitie voltage gives an infinitie amount of energy - of course paying the bill might be tricky!