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
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
Attachments

134.8 KB Views: 6