The process you're fishing for is "polarization", but that's a
misleading description.
Polarization doesn't do anything to change the light waves.
It simply filters out (absorbs, as with a polarizing filter) the
light waves that aren't vibrating in the desired plane, and
allows only those that are to pass.
The intensity of a light beam is always reduced after
polarizing it, because much (most) of the original light
has been removed.
A laser light source may be thought of as an exception,
since everything coming out of the laser is polarized.
In series circuit, Req = R₁ + R₂ + R₃ + ···
In parallel circuit, 
<h3>Q7.</h3>
total resistance in the upper branch = R₂ + R₃ = R₂ + 2


R₂ + 2 = 12
R₂ = 10Ω
<h3>Q8.</h3>


Req = 1.7Ω
Answer:
at the melting point and boiling point
Explanation:
the heat abdorbed is used to break down the forces holding the particles together to allow it to change its state