The voltage across an ideal inductor satisfies the following equation (see Fig. 1 for notations) $$\begin{equation}V(t)=L\frac{dI(t)}{dt}\end{equation}$$ which can be inverted to compute the current as a function of voltage $$\begin{equation}I(t)-I(0)=\frac{1}{L}\int_0^t V(\tau) d\tau\end{equation}$$ If $I(0)=0$, the previous equation becomes $$\begin{equation}I(t)=\frac{1}{L}\int_0^t V(\tau) d\tau\end{equation}$$

Notice that, if the current going through an inductor is constant, the voltage is equal to zero. Therefore, the voltage across inductors is equal to zero in DC circuits.

If the current going through the inductor varies very fast, the voltage can increase significantly. Hence, one should pay special attention when working with inductors because any unintended interruption of the current (for instance by accidentally opening a switch) it can lead to large voltages that can damage the inductor itself or other parts of the circuit.

The instantaneous power absorbed by the inductor is $$\begin{equation}P(t)=I(t)V(t)=L\cdot I(t)\frac{dI(t)}{dt} = \frac{L}{2} \frac{dI^2(t)}{dt} \end{equation}$$

while the energy stored in the inductor is $$\begin{equation}W(t)=\frac{L \left[I(t)\right]^2}{2} \end{equation}$$

The current going through an ideal capacitor satisfies the following equation (see Fig. 2 for notations) $$\begin{equation}I(t) = \frac{dQ(t)}{dt} = C\frac{dV(t)}{dt}\end{equation}$$ where $Q$ is the charge stored by the capacitor. The last equation can be inverted to compute the voltage as a function of current $$\begin{equation}V(t)-V(0)=\frac{1}{C}\int_0^t I(\tau) d\tau\end{equation}$$ If $V(0)=0$, the previous equation becomes $$\begin{equation}V(t)=\frac{1}{C}\int_0^t I(\tau) d\tau\end{equation}$$

Notice that, if the voltage across a capacitor is constant, the current is equal to zero. Therefore, the current going through capacitors is equal to zero in DC circuits.

If the voltage across the capacitor varies very fast, the current can increase significantly. Hence, one should always check if a capacitor is charged or not when using it (even if it is disconnected from a circuit). If the capacitor is charged and one accidentally short-circuit its terminals, can lead to large currents that can damage the capacitor itself.

The extra charge that is stored in the capacitor from $0$ to $t$ is $$Q(t)-Q(0)= \int_0^t I(\tau) d\tau = C \int_0^t \frac{dV}{d\tau} d\tau =C \int_0^t dV = C\left[V(t)-V(0)\right]$$ At $V=0$ it is common to use that $Q=0$ (this value can be regarded a as a reference) and $$\begin{equation}Q(t) = C V(t)\end{equation}$$

The instantaneous power absorbed by the capacitor is $$\begin{equation}P(t)=V(t)I(t)=C\cdot V(t)\frac{dV(t)}{dt} = \frac{C}{2} \frac{dV^2(t)}{dt} \end{equation}$$

while the energy stored in the capacitor is $$\begin{equation}W(t)=\frac{C \left[V(t)\right]^2}{2}=\frac{\left[Q(t)\right]^2}{2C} \end{equation}$$

Table 1 shows the fundametal equations for resistors, inductors, and capacitors.