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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 Req= 3+6 = 9 ohm and V = 12Vtherefore current = 4/3A now power = I*I*R = 16 Wnow electrical energy  = P*T = 16 *  60 (s) = 960 ????????

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).