Resistor 1 has twice the resistance of resistor 2.? The two are connected in series and a potential difference is maintained across the combination. The rate of thermal energy generation in 1 is: A....
Resistor 1 has twice the resistance of resistor 2.?
The two are connected in series and a potential
difference is maintained across the combination. The rate of thermal energy generation in 1 is:
A. the same as that in 2
B. twice that in 2
C. half that in 2
D. four times that in 2
E. one-fourth that in 2
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The data is text tells that
`R_1 =2*R` and `R_2=R`
The equivalent resistance of the two resistors connected in series is
`R_(eq) =R_1+R_2 =2R+R =3R`
As Ohm law states the current is the same in both `R_1` and `R_2` :
`I =U/R_(eq) =U/(3R)`
The power (rate of heat generation) in resistor `R_1` is
`P_1 = I^2*R1 =(U/(3R))^2 *(2R) =(U^2/(9R^2))*2R =(2/9)*U^2/R`
The power in resistor `R_2` is
`P_2 =I^2*R2 =(U/(3R))^2 *R =(U^2/(9R^2))*R = (1/9)*U^2/R`
Therefore
`P_1 =2*P_2`
The correct answer is B) twice that in 2
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