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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
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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 ????????

Ashwin , 10 Years ago
Grade 10
anser 2 Answers
shashi K Sharma

Last Activity: 10 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).

TanayaDineshDhondre

Last Activity: 3 Years ago

Given
T= 1m = 60s
R1=3ohm 
R2= 6ohm
V= 12V
As the resistors are connected in series
Hence, Rs = R1+R2
                    =3ohm +6ohm
                    = 9 ohm
Now, by using Ohm's law,
            V= I into R
             I = V/R
               = 12/9
                = 4/3 A
Hence we know that the current remains same in the series.
In case R1,
             R1= 3ohm 
               I = 4/3 A
Hence, V in R1 = I into R
                           = 4/3 into 3
                           = 4 V
Power = V1 into I
             =   4 into 4/3
              = 16/4 W
Electrical energy consumed= P into T 
                                                   = 16/3 into 60 = 16 into 20 = 320J
In case of R2 
                R2 = 6ohm 
                 I = 4/3 A
 Hence potential difference = I into R2
                                                     4/3 into 6 = 8V
Power= V2 into I 
             = 8 into 4/3 = 32/3W
Electrical energy = P into T 
                                = 32/3 into 60 = 32 into 20 = 640J
 
              

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