Basic circuit question (True/False)

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The discussion revolves around a circuit with a source of EMF and identical light bulbs, focusing on the effects of opening and closing a switch. Key points include that when the switch is opened, the potential difference across bulb 4 does not increase, and the current through bulb 2 actually decreases when the switch is closed. The brightness of bulb 1 is consistently greater than that of bulbs 2 and 4 due to the distribution of current in the circuit. The confusion arises from the relationship between current, resistance, and power output in parallel circuits. Overall, the analysis emphasizes the importance of understanding current flow and resistance changes when switches in circuits are manipulated.
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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

Homework Equations



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?
 
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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.
 
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