A 30 ohm resistor is connected in parallel with a variable resistance R. The parallel combination is then connected in series with a 6 ohm resistor and connected across a 120 V source. Find the minimum value of R if the power taken by R is equal to the power taken by the 6 ohm resistor.

Answer 1

The resistance of the parallel component is 30R/(R+30)

If R' is the equivalent resistance of the circuit,

R' = 1/(1/30 + 1/R) + 6
= 36(R+5)/(R+30)

The current through the 6Ω resistor is
120/R' = (10R+300)/(3R+15)

The voltage drop across R is

[30R/(R+30)]/[36(R+5)/(R+30)]*120 = 100R/(R+5)

So, if the power dissipation is the same, and is i^2 r = v^2/r

[(10R+300)/(3R+15)]^2*6 = [100R/(R+5)]^2/R

R = 15(3-√5) ≈ 11.5Ω