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Physics 12 Online
OpenStudy (anonymous):

In which of the circuits does a current exist to light the bulb? @Abhisar

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

OpenStudy (abhisar):

ohhh !!

OpenStudy (anonymous):

#109

OpenStudy (anonymous):

i say 5

OpenStudy (abhisar):

why ?

OpenStudy (anonymous):

trying to picture the inside of a lamp

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

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 ?

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

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

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

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

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??

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??

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!

OpenStudy (sidsiddhartha):

very good ^_^

OpenStudy (anonymous):

\(\huge\ddot\smile\)

OpenStudy (anonymous):

thank you

OpenStudy (sidsiddhartha):

no problem :) anytime

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