# Basic circuit question (True/False)

## Homework Statement

The circuit shows a source of EMF and some identical light bulbs. The switch, S, is opened at some point. The questions refer to the situation before and after the opening of S. Answer true or false for each statement.

1. When S is opened, the potential difference across bulb 4 increases.
2. When S is opened, the current through bulb 2 decreases.
3. When S is closed, bulb 1 is brighter than bulb 4.
4. When S is closed, bulb 1 is brighter than bulb 2.
5. When S is opened, bulb 4 becomes brighter.

Circuit: http://i.imgur.com/E9e5Y.gif

V=IR
P=I2R

## The Attempt at a Solution

I answered the following:

1. False
2. True
3. False
4. False
5. False

I figured that voltage is constant from the battery, and voltage is the same for each bulb in a parallel circuit, so #1 is False.

Taking away bulb 3 from the circuit means more current for bulb 2 and 3, but it also increases the resistance in the parallel circuit (and thus the total resistance), so the current decreases to compensate. Not sure what the answer would be for #2.

#3 and #4 are the same question as far as I can tell, since bulb 2 and bulb 4 should be identical. I just messed with ratios for the last 3 questions, but I'm not really sure about these.

Any help?

## Answers and Replies

Brightness is proportional to power output, and in this case we care about the amount of current passing through. The current is at a maximum through bulb 1, and this divides into 2*1/2 before the switch is closed and 3*1/3 after the switch is closed.

So the current through bulb 1 is always the greatest of the 4, hence the greatest power (since they are all equal resistances).

But power also depends on resistance, and the resistance is greater in bulb 1. Does this not matter?

I thought you said the bulbs were identical?

Sorry, yea, just confused myself for a second. Any idea for #1, 2 & 5?

1 should be correct.

As for 2, when the switch is closed, the total resistance goes from (1.5)R to (1.333)R. Less resistance means increasing the total current pulled from the voltage source. In fact, you can find the total open-switch current, iopen=V/(1.5R), and the closed-switch current, iclosed=V/(1.333R).

You know that when the switch is open, half the total current goes through bulb 2, and after the switch is closed, 1/3 the total current goes through bulb 2.

So for the currents through bulb 2 before and after switching:

open: ibefore_bulb2=V/(3R) = .333 V/R
closed: iafter_bulb2=V/(4R) = .250 V/R

So the current through bulb 2 decreases when you CLOSE it. Increases when you open it.