Resistors 4Ω,6Ω and 12Ω are connected in parallel, its equivalent resistance (R) is given by R1=41+61+121⇒R=612=2Ω Again R is connected to 1.5 V battery whose internal resistance r=1Ω. Equivalent resistance now, R′=2Ω+1Ω=3Ω Current, Itotal =R′V=31.5=21 A Itotal =21=3x+2x+x=6x⇒x=121 ∴ Current through 4Ω resistor =3x =3×121=41 A Therefore, rate of Joule heating in the 4Ω resistor =I2R=(41)2×4=41=0.25 W