电容器充电

Suppose we have a capacitor of capacitance $C$ and a resistor of resistance $R$ connected to a battery of potential difference $V$. This circuit is called an RC circuit. Suppose that at time $t=0$, the capacitor has zero charge. Let $Q$ be the charge on the capacitor, and let $I=\frac{dQ}{dt}$ be the current. Define $V_C$ and $V_R$ to be the potential differences across the capacitor and resistor, respectively.

Let us first build an intuition for the situation. At time $t=0$, since $Q=0$, we would expect the current to be at its maximum, so that the capacitor can start to charge. However, as time goes on, we expect the current to decrease as the capacitor approaches its maximum charge. Ultimately, after a very long time (we can say as $t\rightarrow\infty$), it seems that $I$ should approach $0$, while $Q$ should approach its maximum value, $CV$.

To analyze the RC circuit mathematically, we use Kirchhoff's Loop Rule, which states that the total potential difference is zero over a complete loop: $$V-\frac{Q}{C}-IR=0$$ $$\frac{dQ}{dt}=\frac{CV-Q}{RC}$$

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

This answer is intuitively pleasing, since it agrees with our earlier notions that $Q=0$ at $t=0$ and $Q\rightarrow CV$ 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 current, we can simply differentiate: $$I=\frac{dQ}{dt}=\frac{V}{R}e^{-t/RC}$$

This also agrees with our intuition, since $I$ starts out at its maximum value and tends toward zero as the capacitor charges and approaches its maximum charge.

Note the constant $RC$. This is called the time constant of the circuit, and it is often denoted $\tau$. You are encouraged to check for yourself that $\tau=RC$ indeed has dimension of time. The time constant tells us how quickly or slowly the capacitor charges to 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: $$Q=0$$ $$V_C=0$$ $$V_R=V$$ $$I=\frac{V}{R}$$ An easy way to remember this is as follows: a capacitor with no charge acts as a short-circuit.

Also note that, as $t\rightarrow\infty$, we have: $$Q=CV$$ $$V_C=V$$ $$V_R=0$$ $$I=0$$ An easy way to remember this is as follows: a fully charged capacitor acts as an open circuit.