Three resistors of 4 , 6 and 12 are connected in parallel and the combination is connected in series with a 1.5 ~V battery of 1 internal resistance. The rate of Joule heating in the 4 resistor is

Step-by-step solution

Resistors 4 , 6 and 12 are connected in parallel, its equivalent resistance (R) is given by 1 R = 1 4 + 1 6 + 1 12 R= 12 6 =2 Again R is connected to 1.5 ~V battery whose internal resistance r=1 . Equivalent resistance now, R^ =2 +1 =3 Current, I_ total = V R^ = 1.5 3 = 1 2 ~A aligned & I_ total = 1 2 =3 x+2 x+x=6 x & x= 1 12 aligned Current through 4 resistor =3 x =3 1 12 = 1 4 ~A Therefore, rate of Joule heating in the 4 resistor =I^2 R= ( 1 4 )^2 4= 1 4 =0.25 ~W