Power in a RCL circuit

Question

I have an ideal RCL series circuit. How can I find the input power that the circuit needs. I know that I can write:

$$V_{in}=V_R+V_C+V_L$$

Answer 1

Well, as you stated in your question we know:

$$ \begin{cases} \text{V}_\text{in}\left(t\right)=\text{V}_\text{R}\left(t\right)+\text{V}_\text{C}\left(t\right)+\text{V}_\text{L}\left(t\right)\\ \\ \text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}\\ \\ \text{I}_\text{C}\left(t\right)=\text{V}_\text{C}'\left(t\right)\cdot\text{C}\\ \\ \text{V}_\text{L}\left(t\right)=\text{I}_\text{L}'\left(t\right)\cdot\text{L}\\ \\ \text{I}_\text{in}\left(t\right)=\text{I}_\text{R}\left(t\right)=\text{I}_\text{C}\left(t\right)=\text{I}_\text{L}\left(t\right) \end{cases}\tag1 $$

So, we can write:

$$\text{V}_\text{in}'\left(t\right)=\text{I}_\text{in}'\left(t\right)\cdot\text{R}+\text{I}_\text{in}\left(t\right)\cdot\frac{1}{\text{C}}+\text{I}_\text{in}''\left(t\right)\cdot\text{L}\tag2$$

And so for the power we know:

$$\text{P}_\text{in}\left(t\right)=\text{V}_\text{in}\left(t\right)\cdot\text{I}_\text{in}\left(t\right)\tag3$$

Answer 2

Evidently, the circuit is a series circuit. Note that we have the following relationships

$$\begin{align} V_R&=RI\\\\ V_L&=L\frac{dI}{dt}\\\\ V_C&=\frac1C \int I\,dt \end{align}$$

The input power is given by $P_{in}=V_{in}I$. Thus,

$$\begin{align} P_{in}&=RI^2+LI\frac{dI}{dt}+\frac1C I\int I\,dt\\\\ &=RI^2+LI\frac{dI}{dt}+C V_C\frac{dV_C}{dt}\\\\ &=RI^2+\frac12 L\frac{dI^2}{dt}+\frac12 C\frac{dV_C^2}{dt}\\\\ &=RI^2+\frac{d}{dt}\left(\frac12LI^2+\frac12CV_C^2\right)\\\\ \end{align}$$

which shows that the input power is given by the sum of the power dissipated by the resistor plus the rate of change in the energy stored in the inductor and capacitor.