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A 30 ohm resistor is connected in parallel with a variable resistance R and the parallel combination is then connected in serie...Asked by Oscar
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.
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Answered by
Steve
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
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
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