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.
Well, that's a shocking question! Let me calculate the minimum value of your resistance and give you a serious answer, just for the watt of it.
First, let's calculate the total resistance of the parallel combination. We can use the formula:
1/Req = 1/R1 + 1/R2 + 1/R3 + ...
In this case, R1 is the 30 ohm resistor and R2 is the variable resistance R. So, we can write:
1/Req = 1/30 + 1/R
Now, to find the minimum value of R, we want the power taken by R to be equal to the power taken by the 6 ohm resistor. The power in a resistor can be calculated using the formula:
P = V^2 / R
Where P is the power, V is the voltage, and R is the resistance. In this case, the voltage is 120 V, so we can write:
P(R) = 120^2 / R
P(6) = 120^2 / 6
To find the minimum value of R, we need to solve the equation P(R) = P(6):
120^2 / R = 120^2 / 6
Now, let's solve for R:
R = (120^2 / 6) / (120^2 / R)
R = 6R / 6
And the answer is...
Drumroll, please...
R = 6 ohms!
I hope that ohmuses you!
To find the minimum value of R, we need to set up an equation based on the condition that the power taken by R is equal to the power taken by the 6 ohm resistor.
We know that power can be calculated using the formula:
P = (V^2) / R
where P is the power, V is the voltage, and R is the resistance.
For the 6 ohm resistor, the power can be calculated as:
P_6 = (120^2) / 6 = 2400 W
For the parallel combination of the 30 ohm resistor and R, the equivalent resistance can be calculated as:
1/Req = 1/30 + 1/R
Simplifying the expression:
Req = (30R) / (30 + R)
The voltage across the parallel combination will be the same as the source voltage, which is 120 V.
Using Ohm's law, we can find the current through the parallel combination:
I = V / Req = 120 / ((30R) / (30 + R)) = (3600 + 120R) / (30R)
The power taken by the parallel combination is:
P_parallel = (I^2) * Req = ((3600 + 120R)^2 / (30R)) * ((30R) / (30 + R))
P_parallel = (3600 + 120R)^2 / (30 + R)
Since we want the power taken by R to be equal to the power taken by the 6 ohm resistor, we have:
P_parallel = P_6
(3600 + 120R)^2 / (30 + R) = 2400
Simplifying the equation:
(3600 + 120R)^2 = 2400 * (30 + R)
1296R^2 + 864000R + 12960000 = 72000 + 2400R
1296R^2 + 840000R + 120000 = 0
We can solve this quadratic equation to find the values of R. However, since we want to find the minimum value of R, we can ignore the negative solution.
Using the quadratic formula:
R = (-b + sqrt(b^2 - 4ac)) / (2a)
R = (-840000 + sqrt(840000^2 - 4*1296*120000)) / (2*1296)
R ≈ 70.3 ohms
Therefore, the minimum value of R is approximately 70.3 ohms.
To find the minimum value of R, we can start by determining the power taken by each resistor.
In a parallel combination of resistors, the total resistance (Rp) is given by:
1/Rp = 1/R1 + 1/R2
In this case, the 30 ohm resistor and the variable resistor R are connected in parallel, so:
1/Rp = 1/30 + 1/R
The total current flowing through the parallel combination is the same as the current flowing through the 6 ohm resistor. Let's call this current I.
Using Ohm's Law, we can calculate the current flowing through the 6 ohm resistor:
I = V/Rtotal
where V is the voltage of the source (120 V) and Rtotal is the total resistance.
Now, let's find the power taken by each resistor.
The power dissipated by a resistor can be calculated using the formula:
P = I^2 * R
For the 6 ohm resistor, the power is:
P6 = I^2 * 6
For the parallel combination, the power is:
PRp = I^2 * Rp
Since we want to find the minimum value of R where the power taken by R is equal to the power taken by the 6 ohm resistor (PRp = P6), we can set up an equation:
I^2 * Rp = I^2 * 6
Simplifying the equation, we can cancel out I^2:
Rp = 6
Now we can substitute Rp into the equation for the parallel combination resistance:
1/Rp = 1/30 + 1/R
1/6 = 1/30 + 1/R
We can now solve for R.
To add fractions, we need a common denominator. In this case, it is 30R:
30R/6 = (30R/30) + (6/30)
After simplifying, we get:
5R = R + 1
Subtracting R from both sides, we get:
4R = 1
Finally, isolating R, we find:
R = 1/4
Therefore, the minimum value of R for which the power taken by R is equal to the power taken by the 6 ohm resistor is 1/4 ohms.
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