Three resistors of  $4\Omega ,6\Omega and12\Omega$ are connected in parallel and the combination is connected in series with a 1.5 V battery of $1\Omega$ internal resistance. The rate of Joule heating in the $12\Omega$  resistor is $\frac{3}{P}$ watt. The value of $P$ is ______

Resistors $4\Omega ,6\Omega and12\Omega$  are connected in parallel, its equivalent resistance (R) is given by
 $\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}\Rightarrow R=\frac{12}{6}=2\Omega$
Again R is connected to 1.5 V battery whose internal resistance  $r=1\Omega$
Equivalent resistance now,
$R\text{'}=2\Omega +1\Omega =3\Omega$ 
Current,  $I_{total}=\frac{V}{R\text{'}}=\frac{1.5}{3}=\frac{1}{2}A$
 $I_{total}=\frac{1}{2}=3x+2x+x=6x\Rightarrow x=\frac{1}{12}$
$\therefore$  Current through $4\Omega$  resistor $=3x$
$=3\times \frac{1}{12}=\frac{1}{4}A$ 
Therefore, rate of Joule heating in the $4\Omega$  resistor
 $=I^{2}R={\left(\frac{1}{4}\right)}^2\times 4=\frac{1}{4}=0.25W$