How to Find the Rate of Change of Resistance

The phrase “rate of change of resistance” needs a variable: resistance might change over time, with temperature, or as a wire’s dimensions change. The derivative depends on which quantity is changing. This article shows how to choose the derivative, calculate it from common resistance models, and interpret its units.

Choose the independent variable

If resistance $R$ changes over time $t$, the instantaneous rate is

$$\frac{dR}{dt}.$$

Its units might be ohms per second, written $\Omega/\text{s}$. If you are instead asking how resistance varies with temperature $T$, use $dR/dT$, with units such as $\Omega/{}^\circ\text{C}$. That temperature derivative is not itself a rate per unit time; to get a time rate, use the chain rule:

$$\frac{dR}{dt}=\frac{dR}{dT}\frac{dT}{dt}.$$

The same rule applies when resistance depends on another changing quantity, such as a wire’s length or cross-sectional area. Before differentiating, identify the variable and the physical model relating it to resistance.

Use the model that describes the situation

For a uniform wire, resistance is modeled by

$$R=\rho\frac{L}{A},$$

where $\rho$ is the material’s resistivity, $L$ is its length, and $A$ is its cross-sectional area. If only the length changes and $\rho$ and $A$ stay constant, then

$$\frac{dR}{dt}=\frac{\rho}{A}\frac{dL}{dt}.$$

If more than one quantity changes, differentiate the full expression. In relative-rate form,

$$\frac{1}{R}\frac{dR}{dt} = \frac{1}{\rho}\frac{d\rho}{dt} +\frac{1}{L}\frac{dL}{dt} -\frac{1}{A}\frac{dA}{dt},$$

assuming $\rho$, $L$, and $A$ are nonzero and differentiable. This form makes the contributions clear: increasing resistivity or length tends to increase resistance, while increasing area tends to decrease it.

For many materials over a limited temperature range, a common approximation is

$$R(T)=R_0\bigl(1+\alpha(T-T_0)\bigr),$$

where $R_0$ is resistance at reference temperature $T_0$ and $\alpha$ is the temperature coefficient of resistance. Differentiating gives $dR/dT=R_0\alpha$, so if temperature changes with time,

$$\frac{dR}{dt}=R_0\alpha\frac{dT}{dt}.$$

This linear approximation is useful over a suitable temperature range; it is not a universal law for every material or condition.

Worked example: resistance warming over time

Suppose a component has resistance $R_0=10\,\Omega$ at $T_0=20^\circ\text{C}$, and its temperature coefficient is $\alpha=0.004/{}^\circ\text{C}$. At a particular moment its temperature is rising at $2^\circ\text{C}/\text{min}$. Under the linear model,

$$\begin{aligned} \frac{dR}{dt} &=R_0\alpha\frac{dT}{dt}\\ &=(10\,\Omega)(0.004/{}^\circ\text{C})(2^\circ\text{C}/\text{min})\\ &=0.08\,\Omega/\text{min}. \end{aligned}$$

At that moment, resistance is increasing by $0.08\,\Omega$ per minute. The temperature units cancel, leaving resistance per time. This is an instantaneous rate; it does not mean resistance rises by exactly that amount during every later minute if the heating rate or model changes.

If voltage and current are given

When voltage $V$ and current $I$ are known and current is nonzero, Ohm’s law gives $R=V/I$. If both vary with time, differentiate the quotient:

$$\frac{dR}{dt} = \frac{I\,dV/dt-V\,dI/dt}{I^2}.$$

For example, if $V=12\,\text{V}$ and $I=2\,\text{A}$ at a given instant, with $dV/dt=0.5\,\text{V}/\text{s}$ and $dI/dt=0.1\,\text{A}/\text{s}$, then

$$\frac{dR}{dt} = \frac{(2)(0.5)-(12)(0.1)}{2^2} =-0.05\,\Omega/\text{s}.$$

The negative sign means the resistance inferred from $V/I$ is decreasing at that instant. If either $V$ or $I$ is not changing, set its derivative to zero in the formula.

When resistance follows a differential equation

Sometimes the model directly specifies how resistance changes, for example

$$\frac{dR}{dt}=kR,$$

where $k$ is a constant. This equation says that the instantaneous rate is proportional to the current resistance. Its solutions are $R(t)=R(0)e^{kt}$, and the rate at time $t$ is $kR(0)e^{kt}$. A differential equation is needed only when the problem gives a rule of this kind or when a physical model leads to one; it is not required for every rate-of-change question.

Common mistakes

• Leaving out “with respect to what.” $dR/dt$ and $dR/dT$ are different quantities and have different units.
• Using the wrong derivative. If temperature changes over time, multiply $dR/dT$ by $dT/dt$ to obtain $dR/dt$.
• Treating a changing quantity as constant. In $R=\rho L/A$, only hold a factor constant if the situation says it is fixed or the model justifies it.
• Dropping units or the sign. Units identify the rate, and a negative derivative means resistance is decreasing in the chosen direction of change.
• Confusing a rate with a total change. A derivative is an instantaneous rate. To find a finite change, use the model at both endpoints or integrate the rate over the interval.

If the problem provides a specific equation or measurements, first state the changing variable, then differentiate that particular model. Without those details, there is no unique numerical rate of change.