Parallel circuits with switches

Apr 2011
24
0
I attached a homework problem with parallel circuits with two switches, and I was hoping someone could lend me a hand with my reasoning here.

I named the switch on the left switch 1 and the switch on the right switch 2. So here is what i came up with, but i am not entirely sure if my answers are correct:

switch 1 open and switch 2 open: only bulb A would light and would thus be the brightest bulb because it is the only one that stays in circuit

switch 1 open and switch 2 closed: only bulb A would light again

switch 1 closed and switch 2 open: bulbs A, B, and C would light, with A being twice as bright as B and C because the same current goes to both parallel branches, where B and C have to split the current that A gets by itself

swithc 1 closed and 2 closed: A would be three times as bright as D, E, and F, and A would still be twice as bright as B and C. So the ranking by brightness would be A, B and C (tied), then D, E, and F (tied)


Am I on the right track here?

Thanks in advance for any help!

Please!

Ray
 

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topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
I attached a homework problem with parallel circuits with two switches, and I was hoping someone could lend me a hand with my reasoning here.

I named the switch on the left switch 1 and the switch on the right switch 2. So here is what i came up with, but i am not entirely sure if my answers are correct:

switch 1 open and switch 2 open: only bulb A would light and would thus be the brightest bulb because it is the only one that stays in circuit

switch 1 open and switch 2 closed: only bulb A would light again

switch 1 closed and switch 2 open: bulbs A, B, and C would light, with A being twice as bright as B and C because the same current goes to both parallel branches, where B and C have to split the current that A gets by itself

swithc 1 closed and 2 closed: A would be three times as bright as D, E, and F, and A would still be twice as bright as B and C. So the ranking by brightness would be A, B and C (tied), then D, E, and F (tied)


Am I on the right track here?

Thanks in advance for any help!

Please!

Ray
You have the right answers ( ;) ), but the wrong reasoning. The potential difference across the A lamp, the B and C lamps, and the D and E and F lamps are all the same. The resistances are correspondingly greater as the number of lamps grows, thus you have less current to go through the branches with larger numbers of lamps and thus the branches with more lamps are dimmer.

-Dan
 
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