Answer
$0.99W; 1.32W; 1.13W$
Work Step by Step
We know that at the junction $I=I_1+I_2$....eq(1)
In the lower closed loop $12V-I(12\Omega)-I_1(65\Omega)=0$
This simplifies to:
$I_1=\frac{12}{65}-\frac{12}{65}I$.....eq(2)
In the upper closed loop
$12V-I(12\Omega)-I_2(55\Omega)=0$
This simplifies to:
$I_2=\frac{12}{55}-\frac{12}{55}I$....eq(3)
We plug in values from eq(2) and eq(3) in eq(1) to obtain:
$I=(\frac{12}{65})-(\frac{12}{65})I+(\frac{12}{65})-(\frac{12}{65})I$
This simplifies to:
$I=0.287A$
We plug in this value in eq(2) to obtain:
$I_1=\frac{12}{65}-\frac{12}{65}(0.287A)$
$\implies I_1=0.132A$
and from eq(1) $I_2=I-I_1$
$I_2=0.287A-0.132A$
$I_2=0.155A$
Now the power dissipated in the $12\Omega$ resistor
$P_{12}=(0.287A)^2(12\Omega)=0.99W$
Power dissipated in $55\Omega$
$P_{55}=(0.155A)^2(55\Omega)=1.32W$
Power dissipated in $65\Omega$
$P_{65}=(0.132A)^2(65\Omega)=1.13W$
