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
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 !!!!!
To solve this problem, we can use the formulas for calculating resistances, currents, and power in a parallel and series circuit.
First, let's determine the equivalent resistance of the parallel circuit. The formula to calculate the equivalent resistance for resistances in parallel is:
1/Req = 1/R1 + 1/R2 + 1/R3 + ...
Given the resistances R1 = 12 Ω, R2 = 18 Ω, and R3 = 36 Ω, we can calculate the equivalent resistance as follows:
1/Req = 1/12 + 1/18 + 1/36
1/Req = (3 + 2 + 1)/36
1/Req = 6/36
1/Req = 1/6
Taking the reciprocal of both sides, we get:
Req = 6 Ω
Now, let's calculate the current flowing through the parallel circuit. The formula to calculate the current flowing through a branch in a parallel circuit is:
I = V/Req
Given the supply voltage V = 60 V and equivalent resistance Req = 6 Ω, we can calculate the current as follows:
I = 60/6
I = 10 A
Since the resistances in the parallel circuit have the same voltage across them, the power dissipated in each resistance is equal. We are given that the power dissipated in the 12 Ω resistance is 36 W.
Using the formula for power dissipation in a resistor:
P = I^2 * R
where P is the power, I is the current, and R is the resistance, we can calculate the current flowing through the 12 Ω resistance:
36 = I^2 * 12
Rearranging the equation, we get:
I^2 = 36/12
I^2 = 3
I = √3
Now, let's calculate the current flowing through the series circuit using the formula:
I_total = I_parallel = √3 A
Since the circuit is in series, the current flowing through each component is the same. Therefore, the current flowing through the fourth resistance is also √3 A.
Lastly, let's determine the value of the fourth resistance. We can use the formula for calculating the resistance in a series circuit:
R_total = R1 + R2 + R3 + R4
Given R1 = 12 Ω, R2 = 18 Ω, R3 = 36 Ω, and R_total = Req + R4, we can substitute the known values:
R_total = 6 + R4
Since the total resistance is equal to the sum of the equivalent resistance and the fourth resistance, we have:
6 + R4 = R_total
Substituting the value of the total resistance R_total = Req + R4 = 6 + R4, we get:
6 + R4 = √3
Solving for R4, we have:
R4 = √3 - 6
So, the value of the fourth resistance is approximately -3.46 Ω.
To find the total power dissipated in the group, we can add up the power dissipated in each of the resistors in the series circuit, using the power formula:
P_total = I_total^2 * R_total
Given I_total = √3 A and R_total = 6 + R4, we can calculate the total power dissipated as follows:
P_total = (√3)^2 * (6 + R4)
P_total = 3 * (6 + √3 - 6)
P_total = 3 * √3
So, the total power dissipated in the group is approximately 9√3 W.