Question

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?

Answer 1

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:

$V = i\cdot R$ (1)

Where:

$V$ - Voltage, measured in volts.

$i$ - Current, measured in amperes.

$R$ - Resistance, measured in ohms.

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

$\frac{dV}{dt} = \frac{di}{dt}\cdot R +i\cdot \frac{dR}{dt}$

$\frac{dV}{dt} = \frac{di}{dt}\cdot R + \frac{V}{R}\cdot \frac{dR}{dt}$ (2)

Where:

$\frac{dV}{dt}$ - Rate of change in voltage, measured in volts per second.

$\frac{di}{dt}$ - Rate of change in current, measured in amperes per second.

$\frac{dR}{dt}$ - Rate of change in resistance, measured in ohms per second.

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

$5\cdot \frac{di}{dt}-1.6 = 0$ (3)

$\frac{di}{dt} = 0.32\,\frac{A}{s}$

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