RL电路方程

Suppose we have an inductor of inductance $L$ and a resistor of resistance $R$ connected in series to a battery of potential difference $V$. This circuit is called an RL circuit. Suppose that at time $t=0$, a switch in the circuit is flipped so as to close the circuit. Define $V_L$ and $V_R$ to be the potential differences across the inductor and resistor, respectively.

Let us first build an intuition for the situation. At time $t=0$, the inductor would fight the change in current, so $I$ would not just immediately jump to $\frac{V}{R}$. Instead, it seems that it will start at $0$ and increase toward $\frac{V}{R}$ as time passes. Ultimately, after a very long time (we can say as $t\rightarrow\infty$), it seems that $I$ should approach its maximum value $\frac{V}{R}$, as if there were no inductor at all. Examining $\frac{dI}{dt}$, it seems that it should initially be at its maximum, so that $I$ can start to increase. Then, as the current approaches its maximum, $\frac{dI}{dt}$ should tend to $0$.

To analyze the RC circuit mathematically, we use Faraday's Law, which states that, over a closed loop $X$: $$\int_{X} \vec{E}\cdot\vec{dl}=-\frac{d\phi}{dt}$$ From the above, we obtain: $$V-IR=-L\frac{dI}{dt}$$ $$\frac{dI}{dt}=\frac{IR-V}{L}$$

This is a standard differential equation, which we can solve quite easily: $$\frac{dI}{IR-V}=\frac{dt}{L}$$ Since $I=0$ at $t=0$, we may integrate as follows: $$\int_{0}^{I}\frac{dI}{IR-V}=\int_{0}^{t}\frac{dt}{L}$$ $$\frac{1}{R}\ln\frac{V}{V-IR}=\frac{t}{L}$$ $$1-I\frac{R}{V}=e^{-\frac{t}{L/R}}$$ $$I=\frac{V}{R}\left(1-e^{-\frac{t}{L/R}}\right)$$

This answer is intuitively pleasing, since it agrees with our earlier notions that $I=0$ at $t=0$ and $I\rightarrow \frac{V}{R}$ as $t\rightarrow\infty$. Now that we have formally worked through the calculations, we can be assured that our intuition is correct.

To find the rate of chnnge of the current, we can simply differentiate: $$\frac{dI}{dt}=\frac{V}{L}e^{-\frac{t}{L/R}}$$

This also agrees with our intuition, since $\frac{dI}{dt}$ starts out at its maximum value and tends toward zero as the current increases and approaches a maximum.

Note the constant $L/R$. This is called the time constant of the circuit, and it is often denoted $\tau$. You are encouraged to check for yourself that $\tau=L/R$ indeed has dimension of time. The time constant tells us how quickly or slowly the current reaches its maximum value.

It's great that we have solved for all of the variables at any time $t$. However, the special times $t=0$ and $t=\infty$ still deserve some special attention.

Note that, at $t=0$, we have: $$I=0$$ $$V_R=0$$ $$V_L=V$$ $$\frac{dI}{dt}=\frac{V}{L}$$ An easy way to remember this is as follows: an inductor initially acts as an open circuit.

Also note that, as $t\rightarrow\infty$, we have: $$I=\frac{V}{R}$$ $$V_R=V$$ $$V_L=0$$ $$\frac{dI}{dt}=0$$ An easy way to remember this is as follows: an inductor acts as a short-circuit after a very long time.