determine the value of fourth resistance and the total power dissipated in the group.

2 b] A circuit consisting of 12Ω, 18 Ω and 36 Ω respectively, joined in parallel, is connected in series with a fourth resistance. The whole is supplied at 60V and it is found that the power dissipated in 12 Ω resistance is 36 W. determine the value of fourth resistance and the total power dissipated in the group.

P_12Ω = 36 W, R=? PT =?
P_12Ω = V²_{12 Ω} / 12
Therefore V²_{12 Ω} = P_12Ω x12 = 36 X12 = 432
V_{12 Ω} = √𝟒𝟑2 =20.78 v

The voltage across RΩ = 60 – 20.78 = 39.2 V
V_RΩ =39.2 V
The current through 12Ω I_{12 Ω} = V12 Ω / 12 [ I = V/R]
= 20.78/12 = 1.73 A
The current through 18Ω I_{18 Ω} = V18 Ω / 18
= 20.78/18 = 1.15 A
The current through 36Ω I_{36 Ω} = V36 Ω / 18
= 20.78/36 = 0.57 A
Therefore the total current I_T = I_{12 Ω} + I_{18 Ω} + I_{36 Ω} = I_RΩ
I_T= I_RΩ = 1.73 +1.15+0.57 = 3.45 A
IRΩ = 3.45 A
Therefore R = V_RΩ / I_RΩ = 39.2 / 3.45
R = 11.36Ω
The total power P_T = V_T I_T = 60 x 3.45 =207W