Question

Two resistances 6ω and 3ω are connected in parallel and this combination is connected in series with a 4ω resistance. this combination is powered by a voltage source of 12 v and zero internal resistance. the ratio of power dissipated between 6ω resistance and 4ω resistance is

Answer 1

two resistors of resistance 6Ω and 3Ω are connected in parallel and this combination is connected in series with 4Ω resistor .
See attachment, equivalent resistance will be :
1/Req = R₁R₂/(R₁ + R₂) + R₃
Here, R₁ = 6Ω , R₂ =3Ω and R₃ = 4Ω
Now, Req = 6Ω
So, current , i = V/Req = 12/6 = 2A
Potential across 6Ω = potential across 3Ω [cause both are in parallel ]
i₁ × R₁ = i₂ × R₂
i₁ × 6 = i₂ × 3 ⇒ 2i₁ = i₂
Means , current 2A divides 2 : 1 ratio in 6Ω and 3Ω resistor .
so, current passing through 6Ω = 4/3 A
Now, power dissipated by 6Ω = i₁²R₁
= (4/3)² × 6 = 16 × 6/9 = 32/3 W

power dissipated by 4Ω = i²R₃
= (2)² × 4 = 16

HENCE, ratio of power dissipated by 6Ω and 4Ω = 32/3/16 = 2 :3

Answer 2

Answer:

3 : 2

Explanation:

    first parallel resistance  1/R p = 1/6 + 1/3

                                ------------      R p = 2 ohm -------

  series resistance R s = 2 + 4

              ----------  R s = 6 ohm  -----------

next thing  is ....current I = V/R

                             I = 12 / 6

                            I = 2 A

power....P = I ^2 x  R

              P1 / P2  =    2^2 x 6 /  2^2 x 4

             P1 : P2 = 3 : 2