A circuit consisting of three resistances 12 Ω, 18 Ω and 36 Ω respectively, joined in
parallel, is connected in series with a fourth resistance. The whole is supplied at 60 V and it
is found that the power dissipated in the 12 Ω resistance is 36 W. Determine the value of the fourth resistance and the total power dissipated in the goup
2 answers
I like this kind of question
conductance 1/resistance
so for the 3
conductance = 1/12 + 1/18 + 1/36 = .08333 + .05555 + .02777
= .1667
Resistance of the three = 1/ conductance = 6 ohms :) LOL
total Resistance = (6 + R)
i total = 60/(6+R)
total current = 60/ (6+R)
what fraction of the total current goes through the 12 ohms ?
i12 = V/12 = .08333 V
i18 = V/18 = .05555 V
i32 = V/32 = .03125 V
sum = .17013 V
fraction through R12 = .08333/.17013 = .49 about half :)
36 watts =i^2 r = (.49 i)^2 (12) = .24 i^2*12 = 2.88 i^2
so i^2 = 36/2.88 = 12.5
i = 3.54 amps through R and through the three
but we know res of || 3 is 6 ohms so
V of ||3 = i R = 3.54 * 6 = 21.2
so
V across R = 60 -21.2 = 38.8
R = 38.8 / 3.54 = 11 ohms
check my arithmetic !!!!!
so for the 3
conductance = 1/12 + 1/18 + 1/36 = .08333 + .05555 + .02777
= .1667
Resistance of the three = 1/ conductance = 6 ohms :) LOL
total Resistance = (6 + R)
i total = 60/(6+R)
total current = 60/ (6+R)
what fraction of the total current goes through the 12 ohms ?
i12 = V/12 = .08333 V
i18 = V/18 = .05555 V
i32 = V/32 = .03125 V
sum = .17013 V
fraction through R12 = .08333/.17013 = .49 about half :)
36 watts =i^2 r = (.49 i)^2 (12) = .24 i^2*12 = 2.88 i^2
so i^2 = 36/2.88 = 12.5
i = 3.54 amps through R and through the three
but we know res of || 3 is 6 ohms so
V of ||3 = i R = 3.54 * 6 = 21.2
so
V across R = 60 -21.2 = 38.8
R = 38.8 / 3.54 = 11 ohms
check my arithmetic !!!!!