Switches changing brightness of bulbs

[AstroJMT42]

Homework Statement

All bulbs are identical, (no information given on resistance so assuming all the same)
All the switches are closed and then S1 is opened, Does bulb 6 get brighter, dimmer or stay the same? Does bulb 4 get brighter, dimmer or stay the same? Explain briefly why.

Homework Equations

None assuming resistance is equal

The Attempt at a Solution

Seeing as there is no info on resistance I am assuming that bulb 6 would actually get brighter as it is the path of least resistance but I am not entirely sure and would like some clarification. Would bulb 4 stay the same? my reasoning is that the circuit splits before bulb 4 so it wouldn't make any difference. Or maybe it would dim as some of the current that was going through bulbs 2 and 3 is no longer splitting and heading towards 4?

 

[WIN]

redraw the circuit and simplify it so that you can see which are series and which are parallel circuit. Then you can calculate it using ohm's law. Simply put a same value for the resistance on each bulb and then a value for volt and you can get the current (I) through each bulb.

 

[Simon Bridge]

What is it that determines the brightness of a bulb?
As WIN suggests - you can just use R for the unknown resistance in your calculations. You should be able to compare the two situations mathematically.

 

[WIN]

higher current through the bulb means brighter bulb ... it depends whether the bulb is in parallel or series connection (which determine the resistance)
just put 100 ohm for every buld and the voltage for the battery as 10 volt ... then simplify and calculate =)

 

[WIN]

it might be a bridge circuit with a missing resistance, not a simple combination of series and parallel.

 

[CWatters]

You don't actually need to do any calculations to answer the questions.

The voltage across bulb 6 can be determined by visual inspection of the circuit alone.

Bulb 4 is slightly harder but can be solved by thinking about the symmetry of the circuit. Not quite sure how to best explain it but perhaps think of resistors or bulbs as "pulling the voltage on a node in a particular direction" (eg either towards the +ve or -ve terminals of the voltage source).

 

[AstroJMT42]

So with bulb 6: In a parallel circuit the amps are shared and the voltage is equal. with S1 shut the current is now split only twice as apposed to 3 times, as the voltage is the same the bulb should be brighter because there are more amps?

Bulb 4 is still perplexing me, most likely because it is in series with bulb 1 but parallel to bulb 5, which is a scenario we didn't cover in class, this is the point in particular that I'm just clearly not getting my head around.

Thanks for the help guys, electronics is by far and away my weakness when it comes to physics.

 

[WIN]

So with bulb 6: In a parallel circuit the amps are shared and the voltage is equal. with S1 shut the current is now split only twice as apposed to 3 times, as the voltage is the same the bulb should be brighter because there are more amps?

Bulb 4 is still perplexing me, most likely because it is in series with bulb 1 but parallel to bulb 5, which is a scenario we didn't cover in class, this is the point in particular that I'm just clearly not getting my head around.

Thanks for the help guys, electronics is by far and away my weakness when it comes to physics.

try google for bridge connection

 

[CWatters]

So with bulb 6: In a parallel circuit the amps are shared and the voltage is equal. with S1 shut the current is now split only twice as apposed to 3 times, as the voltage is the same the bulb should be brighter because there are more amps?

Almost.

The voltage across bulb 6 is the same with S1 open or closed. The total current coming from the battery will change (not constant) so your approach is incorrect.

Hint: The resistance of bulb 6 is constant as well as the voltage.

 

[AstroJMT42]

oh! So if the voltage and the resistance are the same, then using V=IR the current must be the same, therefore the brightness doesn't change for bulb 6?

I'm still not sure whether or not 4 changes though, it is still in series with bulb and the circuit splits before it gets to 5, so would it make no difference?

 

[AstroJMT42]

I just built the circuit using an online tool and the brightness of 6 indeed does not change, I think I understand why now. Thanks for the hints.
It also shows that bulb 4 should actually dim, as its flowing from + to - does this mean the bulb dims because it is no longer getting the extra current flow that would go through bulbs 2, 3 and then split to go through 4?

This must be interesting for you guys, a bit like watching a monkey trying to work out how to crack a nut haha.

 

[CWatters]

oh! So if the voltage and the resistance are the same, then using V=IR the current must be the same, therefore the brightness doesn't change for bulb 6?

Correct.

I'm still not sure whether or not 4 changes though, it is still in series with bulb and the circuit splits before it gets to 5, so would it make no difference?

I can't give you the answer but here are some hints to help you along..

Earlier we determined that the brightness of bulb 6 depends on the voltage across it. The same will apply to bulb 4 so our objective is to find the voltage across bulb 4 (or how the voltage on bulb 4 changes). The voltage across bulb 4 will be the same as the voltage across bulb 5 or the combination of bulb 4 and 5. That's because bulb 4 and 5 are connected to the same nodes (they are in parallel).

First off you can delete bulb 6 and it's switch because that doesn't effect the current flowing in the rest of the circuit.

Then using the rules for series and parallel resistors simplify the circuit as shown in this diagram. If it helps, replace all bulbs with a 1R resistor.

Repeat with S1 Open. You will have two circuits that you can compare and that maybe is sufficient to work out if the voltage V4 increases, decreases or stays the same.