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Suppose that the voltage across the resistor is held constant at 40 volts. If the resistance is steadily decreasing at a rate of 0.2 ohms per second, at what rate is the current changing at the moment that the resistor reaches 5 ohms?

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Answer:

The current is increasing at a rate of 0.32 ampere per second.

Explanation:

The voltage of the resistor is modelled after Ohm's Law, which states that voltage is directly proportional to current:

[tex]V = i\cdot R[/tex] (1)

Where:

[tex]V[/tex] - Voltage, measured in volts.

[tex]i[/tex] - Current, measured in amperes.

[tex]R[/tex] - Resistance, measured in ohms.

An expression for the rate of change in voltage is found by Differential Calculus:

[tex]\frac{dV}{dt} = \frac{di}{dt}\cdot R +i\cdot \frac{dR}{dt}[/tex]

[tex]\frac{dV}{dt} = \frac{di}{dt}\cdot R + \frac{V}{R}\cdot \frac{dR}{dt}[/tex] (2)

Where:

[tex]\frac{dV}{dt}[/tex] - Rate of change in voltage, measured in volts per second.

[tex]\frac{di}{dt}[/tex] - Rate of change in current, measured in amperes per second.

[tex]\frac{dR}{dt}[/tex] - Rate of change in resistance, measured in ohms per second.

If we know that [tex]\frac{dV}{dt} = 0\,\frac{V}{s}[/tex], [tex]R = 5\,\Omega[/tex], [tex]V = 40\,\Omega[/tex] and [tex]\frac{dR}{dt} = -0.2\,\frac{\Omega}{s}[/tex], then the rate of change in current is:

[tex]5\cdot \frac{di}{dt}-1.6 = 0[/tex] (3)

[tex]\frac{di}{dt} = 0.32\,\frac{A}{s}[/tex]

The current is increasing at a rate of 0.32 ampere per second.

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