**This post describes how to calculate power and energy in RC circuit. Energy consumption and power dissipation are very important characteristics of the digital circuit. **

Let’s consider the simple RC circuit with the voltage source as depicted below.

From the previous posts we know that power delivered to a circuit element is $p\left(t\right)=v\left(t\right)i\left(t\right)$. Resistor and capacitor perform different functions in terms of the power in the circuit: resistor – dissipates energy, and capacitor – stores energy.

So the instantaneous power from the source is $p\left(t\right)=Vi\left(t\right)$. Current here is $i\left(t\right)=\frac{V\u2013{v}_{C}\left(t\right)}{R}$. We already know that for this circuit capacitor voltage is ${v}_{C}\left(t\right)=V(1\u2013{e}^{\u2013\frac{t}{RC}})$. Then we have for power $p\left(t\right)=\frac{{V}^{2}}{R}\u2013\frac{{V}^{2}}{R}(1\u2013{e}^{\u2013\frac{t}{RC}})=\frac{{V}^{2}}{R}{e}^{\u2013\frac{t}{RC}}$.

Then we can find that energy supplied by the source $\omega ={\int}_{0}^{T}p\left(t\right)dt$ if $T=\infty $, then supplied energy is $\omega =C{V}^{2}$. Energy stored by capacitor is ${\omega}_{C}\left(t\right)=C\frac{{V}^{2}\left(t\right)}{2}=\frac{C{V}^{2}}{2}$. Instantaneous power on the resistor is $p\left(t\right)=Ri{\left(t\right)}^{2}=\frac{{V}^{2}}{R}{e}^{\u2013\frac{2t}{RC}}$. In order to find the energy dissipated by the resistor is $\omega ={\int}_{0}^{\infty}\frac{{V}^{2}}{R}{e}^{\u2013\frac{2t}{RC}}dt=\frac{C{V}^{2}}{2}$.

Let’s consider the circuit with the switch, that gives the step signal, depicted below.

So here we have two situations here – when capacitor is charges and when capacitors is discharges. It charges from the source when the switch is closed, while resistors ${R}_{1}$ and ${R}_{2}$ are dissipating energy. When the switch is opened, capacitor discharges via the resistor ${R}_{2}$.

The average power is the total amount of energy dissipated during certain interval of time, divided by the length of the time interval $T$, i.e $p=\frac{\omega}{T}$. Where $\omega \left(t\right)={\int}_{0}^{T}p\left(t\right)dt$.

When the switch is on, we can transform the circuit using the Thenevin theorem, we have the following equivalent circuit:

Here ${V}_{TH}=\frac{V{R}_{2}}{{R}_{1}+{R}_{2}}$, and ${R}_{TH}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$. As we considered before the total solution in this case will be the sum of the homogeneous solution and particular solution.

The homogeneous solution here is ${v}_{C}\left(t\right)={V}_{TH}(1\u2013{e}^{\u2013\frac{t}{C{R}_{TH}}})$. The total power dissipated by the resistors will be a sum of power dissipated on resistor ${R}_{1}$ and ${R}_{2}$. In this case we have $p\left(t\right)=\frac{{(V\u2013v}^{}}{}$_{C}

Using the formula for power we can find energy dissipated in the circuit during period of time $0\to {T}_{1}$ is ${\omega}_{0\to T1}={\int}_{0}^{{T}_{1}}(\frac{{\left[V\u2013{V}_{TH}(1\u2013{e}^{\u2013\frac{t}{C{R}_{TH}}})\right]}^{2}}{{R}_{1}}+\frac{{\left[{V}_{TH}(1\u2013{e}^{\u2013{\displaystyle \frac{t}{C{R}_{TH}}}})\right]}^{2}}{{R}_{2}})dt$. After simplification and rearrangement we have ${\omega}_{0\to {T}_{1}}=\frac{{V}^{2}}{{R}_{1}+{R}_{2}}{T}_{1}+\frac{{{V}^{2}}_{TH}C}{2}$.

During the time interval ${T}_{1}\to {T}_{2}$ the switch of the circuit is opens, capacitor discharging and resistor ${R}_{2}$ dissipates energy. Right at the start voltage of the capacitor is ${V}_{TH}$. Capacitor voltage will change with time by the formula ${v}_{C}\left(t\right)={V}_{TH}{e}^{\u2013\frac{t}{{R}_{2}C}}$.

Instantaneous power dissipated in the circuit $p\left(t\right)={v}_{C}\frac{{v}_{c}}{{R}_{2}}=\frac{{{V}_{TH}}^{2}}{{R}_{2}}{e}^{\u2013\frac{2t}{{R}_{2}C}}$, energy here is ${\omega}_{{T}_{1}\to {T}_{2}}={\int}_{{T}_{1}}^{{T}_{2}}p\left(t\right)dt=\frac{C{{V}^{2}}_{TH}}{2}(1\u2013{e}^{\u2013\frac{2{T}_{2}}{C{R}_{2}}})\approx \frac{C{{v}^{2}}_{TH}}{2}$.

The total dissipated energy is the sum of $\omega ={\omega}_{0\to {T}_{1}}+{\omega}_{{T}_{1}\to {T}_{2}}=\frac{{V}^{2}}{{R}_{1}+{R}_{2}}{T}_{1}+{{v}^{2}}_{TH}$. The average dissipated power during period of time $T={T}_{1}+{T}_{2}$ is $\overline{p}=\frac{\omega}{T}=\frac{{V}^{2}}{{R}_{1}+{R}_{2}}\frac{{T}_{1}}{T}+\frac{{{v}^{2}}_{TH}}{T}$.

The average power can be divided to static and dynamic power: ${p}_{static}=\frac{{V}^{2}}{({R}_{1}+{R}_{2})}\frac{{T}_{1}}{T}$, ${p}_{dynamic}=\frac{C{{v}^{2}}_{TH}}{T}$.

Educational content can also be reached via Reddit community **r/ElectronicsEasy**.