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.

How to calculate power and energy in RC circuit

From the previous posts we know that power delivered to a circuit element is p(t)=v(t)i(t). 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(t)=Vi(t).  Current here is i(t)=VvC(t)R.  We already know that for this circuit capacitor voltage is vC(t)=V(1etRC). Then we have for power p(t)=V2RV2R(1etRC)=V2RetRC.

Then we can find that energy supplied by the source ω=0Tp(t)dt if T=, then supplied energy is ω=CV2. Energy stored by capacitor is ωC(t)=CV2(t)2=CV22. Instantaneous power on the resistor is p(t)=Ri(t)2=V2Re2tRC. In order to find the energy dissipated by the resistor is ω=0V2Re2tRCdt=CV22.

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 R1 and R2 are dissipating energy. When the switch is opened, capacitor discharges via the resistor R2.

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=ωT. Where ω(t)=0Tp(t)dt.

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

 

Here VTH=VR2R1+R2, and RTH=R1R2R1+R2. 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 vC(t)=VTH(1etCRTH). The total power dissipated by the resistors will be a sum of power dissipated on resistor R1 and R2.  In this case we have p(t)=(VvC)2R1+v2CR2, where vC=VTH(1etCRTH).

Using the formula for power we can find energy dissipated in the circuit during period of time 0T1 is ω0T1=0T1(VVTH(1etCRTH)2R1+VTH(1etCRTH)2R2)dt.  After simplification and rearrangement we have ω0T1=V2R1+R2T1+V2THC2.

During the time interval T1T2  the switch of the circuit is opens, capacitor discharging and resistor R2 dissipates energy. Right at the start voltage of the capacitor is VTH. Capacitor voltage will change with time by the formula vC(t)=VTHetR2C.

Instantaneous power dissipated in the circuit p(t)=vCvcR2=VTH2R2e2tR2C, energy here is ωT1T2=T1T2p(t)dt=CV2TH2(1e2T2CR2)Cv2TH2.

The total dissipated energy is the sum of ω=ω0T1+ωT1T2=V2R1+R2T1+v2TH. The average dissipated power during period of time T=T1+T2 is p¯=ωT=V2R1+R2T1T+v2THT.

The average power can be divided to static and dynamic power: pstatic=V2(R1+R2)T1Tpdynamic=Cv2THT.

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