Question

Three resistors are connected in series to a 42.1 V battery: R1 = 2.96 Ω, R2 = 7.48 Ω and R3 = 7.42 Ω. Now, as R3 heats up, its resistance increases according to: fraction numerator d R subscript 3 over denominator d t end fraction = 0.552 Ω/s Assume the resistance of the other resistors remain constant. Find fraction numerator d P over denominator d t end fraction , the rate at which the power pulled from the battery is changing in W/s, at the instant R3 = 91.7 Ω. The sign shows whether the power is increasing or decreasing.

Answer 1

Answer:

dP/dt = 26.12 W/s

Explanation:

First, we need to find the value of dt at the instant when R₃ becomes 91.7 Ω. Therefore, we use:

dR₃/dt = 0.552 Ω/s

where,

dR₃ = Change in value of resistance 3 = 91.7 Ω - 7.42 Ω = 84.28 Ω

dt = time interval = ?

Therefore,

84.28 Ω = (0.552 Ω/s)(dt)

dt = (84.28 Ω)/(0.552 Ω/s)

dt = 152.68 s

Now, we find change in power (dP):

dP = V(R₁ + R₂ + dR₃)

dP = (42.1 V)(2.96 Ω + 7.48 Ω + 84.28 Ω)

dP = 3987.71 W

Dividing by dt:

dP/dt = 3987.71 W/152.68 s

dP/dt = 26.12 W/s