What resistance must be connected in parallel to 60 ohms to reduce the combined resistance to 15 ohms?
What resistance must be connected in parallel to a 60 ohm resistor to reduce the combined resistance to 15 ohm
To find the resistance that must be connected in parallel to 60 ohms to reduce the combined resistance to 15 ohms, we can use the formula for resistances in parallel:
1/Req = 1/R1 + 1/R2
Where:
- Req is the combined resistance
- R1 is the resistance that is already connected (60 ohms in this case)
- R2 is the resistance that needs to be connected in parallel (unknown)
Let's substitute the given values into the formula:
1/15 = 1/60 + 1/R2
Now, let's solve for R2:
1/R2 = 1/15 - 1/60
1/R2 = (4/60) - (1/60) = 3/60 = 1/20
To remove the fraction, take the reciprocal on both sides:
R2 = 20 ohms
Therefore, a resistance of 20 ohms should be connected in parallel to 60 ohms to reduce the combined resistance to 15 ohms.
To find the resistance that must be connected in parallel to reduce the combined resistance to 15 ohms, we need to use the formula for the combined resistance of two resistors connected in parallel:
1/(R_total) = 1/(R1) + 1/(R2)
Where R_total is the combined resistance, R1 is the resistance we already have (60 ohms), and R2 is the resistance we need to find.
Plugging in the known values:
1/(15 ohms) = 1/(60 ohms) + 1/(R2)
To simplify the equation, we can find the common denominator on the right side of the equation:
1/(15 ohms) = (R2 + 60 ohms)/(60 ohms * R2)
Now, we can cross-multiply:
(R2 + 60 ohms) * (15 ohms * R2) = (60 ohms * R2)
Expanding:
15 ohms * R2^2 + 900 ohms * R2 = 60 ohms * R2
Rearranging the equation:
15 ohms * R2^2 + (900 ohms - 60 ohms) * R2 = 0
Simplifying further:
15 ohms * R2^2 + 840 ohms * R2 = 0
Now we have a quadratic equation. To solve for R2, we can use the quadratic formula:
R2 = (-b ± √(b^2 - 4ac))/(2a)
Where a = 15 ohms, b = 840 ohms, and c = 0.
Plugging in the values:
R2 = (-840 ohms ± √((840 ohms)^2 - 4 * 15 ohms * 0))/(2 * 15 ohms)
Simplifying:
R2 = (-840 ohms ± √(705,600 ohms^2))/(30 ohms)
Calculating the square root of 705,600:
R2 = (-840 ohms ± 840 ohms)/(30 ohms)
Simplifying further:
R2 = (-840 ohms ± 840 ohms)/(30 ohms)
We have two possible solutions:
R2 = (840 ohms - 840 ohms)/(30 ohms) = 0 ohms
R2 = (-840 ohms - 840 ohms)/(30 ohms) = -56 ohms
Since resistance cannot be negative, the only possible solution is R2 = 0 ohms.
Therefore, to reduce the combined resistance to 15 ohms, a resistance of 0 ohms must be connected in parallel to the existing 60 ohms resistor.
15 = 60R/(60+R)
900 + 15 R = 60 R
45 R = 900