Part of a circuit has two resistors connected in parallel as shown in the figure. R1 is a constant resistor of 10 ohm, while R2 is a variable resistor with resistance that varies at a rate of 2 ohms per minute. The total, or effective, resistance, R, provided by this circuit is given by:
1/R= 1/R1+1/R2

At what rate does the effective resistance change when R2 is 100 ohms? Round off your answer to four decimal places.

164.6723

Step-by-step explanation:

i could be wrong and i apologize if i am. i am new to this topic:)

0.0165

Step-by-step explanation:

R = R1 R2/(R1+R2)

R1 is constant

dR = R1 [(R1+R2) dR2- R2 dR2]/ (R1+R2)^2

dR/dt = R1 dR2/dt [R1]/(R1+R2)^2

dR/dt = 100(2) /(110)^2

a

step-by-step explanation:

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt