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Two coils of resistance 3 Ohm and 6 Ohm . are connected in series across a battery of emf 12v . find electrical energy consumed in 1 min in each resistance when these are connected in series? ans – 640J ----------------------------------------------------------------- i tried it myself but i am getting wrong answer can you please tell where i am making mistake. so R eq = 3+6 = 9 ohm and V = 12V therefore current = 4/3A now power = I*I*R = 16 W now electrical energy = P*T = 16 * 60 (s) = 960 ????????

Two coils of resistance 3 Ohm and 6 Ohm . are connected in series across a battery of emf 12v . find electrical energy consumed in 1 min in each resistance when these are connected in series?
ans – 640J
-----------------------------------------------------------------
i tried it myself but i am getting wrong answer can you please tell where i am making mistake.
 
so Req= 3+6 = 9 ohm 
and V = 12V
 
therefore current = 4/3A 
now power = I*I*R = 16 W
now electrical energy  = P*T = 16 *  60 (s) = 960 ????????

Grade:10

1 Answers

shashi K Sharma
46 Points
6 years ago
Dear Student
 
There is something missing in the answer by book because you were to calculate energy in each resistance which is not same in the two resistors. The complete answer should be 640J in 6ohm and 320J in 3ohm resistor making up a total of 960J and that is waht you are getting. so you are correct and answer is incomplete in the book(i assume so).

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