answer question 69,70
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hey mate !!!
here is your answer !!!
potential difference is => 20 v
resistance => 20 ohm .
v = I R =>
20V = I *20ohm
=> I = 1 ampare
Heat generated in 1 sec =>
H= I ² *R *T
=> 1² *20 * 1 sec
=> 20 j
hence Option number D is correct !!!
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now , your second question :-
AB = 1/ 3 Ohm + 1/3 ohm + 1/3 ohm + 3 ohm
=> 10 ohm
hence , option number B is correct !!!
hope it helps you dear !!!
thanks
here is your answer !!!
potential difference is => 20 v
resistance => 20 ohm .
v = I R =>
20V = I *20ohm
=> I = 1 ampare
Heat generated in 1 sec =>
H= I ² *R *T
=> 1² *20 * 1 sec
=> 20 j
hence Option number D is correct !!!
====================================
now , your second question :-
AB = 1/ 3 Ohm + 1/3 ohm + 1/3 ohm + 3 ohm
=> 10 ohm
hence , option number B is correct !!!
hope it helps you dear !!!
thanks
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Yuichiro13:
Check this out :) Maybe you don't even have the idea of solving the 70 question
Answered by
1
69 )
-> POTENTIAL DIFFERENCE IN A PARALLEL CIRCUIT DOESN'T CHANGE.
=> Total Current = ( V / R ) = ( V / ( R / 4 ) ) = ( 4V / R )
=> Option C is correct !
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70 )
-> Simplifying the Circuit further, by naming the potentials :
---> The potential at the circumference is all same -> " B "
---> The potential at the center is -> "C"
---> The final potential -> "A"
Check out the given Simplification to further solve the problem :
[tex]R_{eq} = [ \frac{1}{3R} + \frac{1}{3R} + \frac{1}{3R} ]^{-1} + R \\ \\ =\ \textgreater \ R_{eq} = R + R = 2R[/tex]
--> Option B is correct !
-> POTENTIAL DIFFERENCE IN A PARALLEL CIRCUIT DOESN'T CHANGE.
=> Total Current = ( V / R ) = ( V / ( R / 4 ) ) = ( 4V / R )
=> Option C is correct !
_____________________________________________________________
_____________________________________________________________
70 )
-> Simplifying the Circuit further, by naming the potentials :
---> The potential at the circumference is all same -> " B "
---> The potential at the center is -> "C"
---> The final potential -> "A"
Check out the given Simplification to further solve the problem :
[tex]R_{eq} = [ \frac{1}{3R} + \frac{1}{3R} + \frac{1}{3R} ]^{-1} + R \\ \\ =\ \textgreater \ R_{eq} = R + R = 2R[/tex]
--> Option B is correct !
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