In which of the circuits does a current exist to light the bulb?
@Abhisar
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OpenStudy (anonymous):
i have no idea
OpenStudy (abhisar):
It seems that your question has a few options. Could you post them too?
We can't help you with questions that provide insufficient information.
OpenStudy (anonymous):
hold on
OpenStudy (abhisar):
???
OpenStudy (anonymous):
scanning picture
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OpenStudy (abhisar):
ohhh !!
OpenStudy (anonymous):
#109
OpenStudy (anonymous):
i say 5
OpenStudy (abhisar):
why ?
OpenStudy (anonymous):
trying to picture the inside of a lamp
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OpenStudy (anonymous):
it's not 5
OpenStudy (anonymous):
no it's 4
OpenStudy (abhisar):
For lighting up a bulb u need the two terminals of the bulb connected to the two terminals of the battery
OpenStudy (abhisar):
Also the two terminals of the battery should not be connected directly to each other
OpenStudy (anonymous):
so none of them
are correct
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OpenStudy (abhisar):
Now ur initial guess was correct
OpenStudy (abhisar):
* no
OpenStudy (anonymous):
i was correct?
OpenStudy (abhisar):
yup
OpenStudy (abhisar):
btw can i ask the name of the book ?
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OpenStudy (anonymous):
ok how bout 110 on scan i believe it goes with picture
OpenStudy (anonymous):
Conceptual Physical Science
OpenStudy (abhisar):
inflow of the current is always equal to the outflow
OpenStudy (anonymous):
and 119
OpenStudy (sidsiddhartha):
for the bulb to glow it have to be connected in parallel to the battery
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OpenStudy (anonymous):
y?
OpenStudy (sidsiddhartha):
in 5 it is connected in parallel
OpenStudy (anonymous):
i'm talking about 119 on the scan
OpenStudy (sidsiddhartha):
oops i was talking about 109
OpenStudy (anonymous):
it's ok
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OpenStudy (sidsiddhartha):
have u solved 109 yet?
OpenStudy (anonymous):
yes, and 110
OpenStudy (anonymous):
need help with 119
OpenStudy (sidsiddhartha):
all right
power= Volts * Amps =\[\frac{ v^2 }{ R }\]
do u know this relation
OpenStudy (anonymous):
no
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OpenStudy (sidsiddhartha):
hmm all right do u know ohm's law?
OpenStudy (anonymous):
no
OpenStudy (sidsiddhartha):
ok so lets solve this conceptually
OpenStudy (sidsiddhartha):
look c is directly parallel with the battery is'nt it?
OpenStudy (anonymous):
yes
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OpenStudy (sidsiddhartha):
so it will draw more current than A and B so it will grow brighter than A and B okay with this?
OpenStudy (anonymous):
y?
OpenStudy (sidsiddhartha):
because its in direct parallel combination it will have a less resistence
OpenStudy (anonymous):
ok
OpenStudy (sidsiddhartha):
and look A ans B those two bulbs are in series combination ok??
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OpenStudy (anonymous):
right
OpenStudy (sidsiddhartha):
and if total resistence of the combination is 2R
OpenStudy (sidsiddhartha):
then A and B both will have resistence R ok??
OpenStudy (anonymous):
ok
OpenStudy (sidsiddhartha):
now they have same resistence
so same current will flow through them ok??
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OpenStudy (anonymous):
so if lightbulb c is unscrewed then a & b will not work?
and if A if unscrew b will not work
OpenStudy (sidsiddhartha):
now so the answer of the first question is
C will glow brightest and A and B will glow with same brightness
and 2nd answer will be--
C will draw the most current
and coming to the third part-----
OpenStudy (sidsiddhartha):
if C is unscrewed then simply C will go out ok?
OpenStudy (sidsiddhartha):
and if A is unscrewed then total combination that is A and B will go out but C will remain as it was
goi it ??
OpenStudy (anonymous):
yep got it!
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