If two resistors R1 and R2 are connected in parallel, then the total resistance R measured in ohms is given by
1/R=1/R1+1/R2
If the resistance of R1 and R2 are increasing at 330 and 660 ohms per second respectively, then the rate of increase of R when R1=5000 and R2=5000 is how many ohms per second?
Just plug and chug
1/R = 1/R1 + 1/R2
1/R = 1/5000 + 1/5000 = 2/5000
R = 2500
-1/R^2 R' = -1/R1^2 R1' - 1/R2^2 R2'
-1/625E4 R' = -1/25E6 * 330 - 1/25E6 * 660
1/625E4 R' = 990/25E6
R' = 990*625E4/25E6 = 990*625/2500
R' = 247.5 ω/s
To find the rate of increase of R when R1 = 5000 ohms and R2 = 5000 ohms, we can use the given equation 1/R = 1/R1 + 1/R2.
First, let's find the initial total resistance R when R1 = 5000 ohms and R2 = 5000 ohms.
Substituting the values into the equation, we get:
1/R = 1/5000 + 1/5000
1/R = 2/5000
To find R, we take the reciprocal of both sides:
R = 5000/2
R = 2500 ohms
Now, let's find the derivatives of R, R1, and R2 with respect to time t.
dR/dt = d(1/R)/dt = -1/R^2 * dR/dt
dR1/dt = 330 ohms/second
dR2/dt = 660 ohms/second
Now, we can substitute these values into the equation and solve for dR/dt when R = 2500 ohms:
dR/dt = -1/(2500^2) * (330 + 660)
dR/dt = -1/6250000 * 990
dR/dt ≈ -0.0001584 ohms/second
Therefore, the rate of increase of R when R1 = 5000 ohms and R2 = 5000 ohms is approximately -0.0001584 ohms/second. Note that the negative sign indicates that the resistance is decreasing over time.