Energy Dissipation/Storage in R, C, and L

The electric power associated with an element (R, C, or L) is the product of the voltage $v(t)$ across and current $i(t)$ through the element:

$$p(t)=\frac{dw}{dt}=\frac{dw}{dq}\;\frac{dq}{dt}=v(t)\,i(t)$$ (37)

Integrating $p(t)$ over a time period $T$, we get the energy

$$w=\int_0^T p(t)\; dt=\int_0^T v(t)\; i(t)\; dt$$ (38)

Depending on its sign, the energy can be either consumed (dissipated, converted to heat) if $w>0$, or stored in the element if $w<0$. We consider specifically the energy dissipation/storage in each of the three types of elements $R$, $C$, and $L$.

• Energy dissipated by resistor $R$

  When a voltage $v(t)$ is applied across $R$, the current through it is $i(t)=v(t)/R$, power consumption is

  $$p(t)=v(t)\;i(t)=\frac{v^2(t)}{R}=i^2(t)R$$ (39)

  The energy dissipated during time period $0 \sim T$ is

  $$w=\int_0^T p(t) dt=\int_0^T v(t)\; i(t) dt =\frac{1}{R}\;\int_0^T v^2(t)\; dt =R\;\int_0^T i^2(t)\; dt$$ (40)

  This energy is converted irreversibly from electrical energy to heat. The rate of dissipation is $p(t)=d\,w/dt$.
  • When a DC voltage $v(t)=V$ is applied across $R$, the current through it is $I=V/R$, and the power consumed by the resistor is

    $$P=VI=\frac{V^2}{R}=I^2R$$ (41)

    The energy dissipation during time period $0 \sim T$ is

    $$w=\int_0^T P\; dt=PT=VIT=\frac{V^2}{R}\,T=I^2RT$$ (42)

  • When a sinusoidal voltage $v(t)=V_p \sin(\omega t)$ is applied across $R$, the current through it is $i(t)=v(t)/R=V_p\sin(\omega t)/R$, and the energy dissipated in time period $T=1/f=2\pi/\omega$ is:

    $$w=\frac{1}{R}\int_0^T v^2(t) dt =\frac{V_p^2}{R}\int_0^T \sin^2(\omega t) dt =\frac{V_p^2}{2R}\int_0^T [1-\cos(2\omega t)] dt=\frac{V_p^2}{2R}\,T$$ (43)

    We define the effective or root-mean-square (RMS) $V_{rms}$ as the equivalent DC voltage needed for $R$ to dissipate the same amount of energy as $v(t)=V_p \sin(\omega t)$:

    $$\frac{{\bf V}^2_{rms}}{R}T=\frac{V_p^2}{2R}T=\frac{1}{R}\int_0^T v^2(t)dt$$ (44)

    Solving for $V_{rms}$ we get

    $$V_{rms}=\sqrt{\frac{1}{T}\int_0^T v^2(t) dt }=\frac{V_p}{\sqrt{2}}=0.707\,V_p$$ (45)

    i.e., the same amount of energy is dissipated by resistor $R$ when either a sinusoidal voltage with peak amplitude $V_p$ or a DC voltage $V_{rms}=V_p/\sqrt{2}$ is applied across it.

    

    Some useful trigonometric identities:

    $$\sin\alpha \cos\alpha=\frac{1}{2} \sin{2\alpha},\;\;\; \sin^2\alpha =\frac{1}{2}[1-\cos(2\alpha)],\;\;\; \cos^2\alpha =\frac{1}{2}[1+\cos(2\alpha)]$$ (46)

    The average of a time varying current $i(t)$ is the value of a DC (direct current) current $I_{av}$ that in period $T$ would transfer the same charge $Q$:

    $$I_{av}T=Q=\int_0^T i(t) dt,\;\;\;\; \;\;\;\; I_{av}=\frac{1}{T}\int_0^T i(t) dt$$ (47)

    Similarly, the average voltage is defined as:

    $$V_{av}=\frac{1}{T}\int_0^T v(t) dt$$ (48)

    If the current/voltage is sinusoidal

    $$i(t)=I_p\,\sin(\omega t)=I_p\,\sin(2\pi ft)=I_p\,\sin(2\pi t/T)$$ (49)

    The average over the complete cycle $T=1/f$ is always zero (the charge transferred during the first half is the opposite to that transferred in the second). However, if we consider the half-cycle over $T/2$, the average is:
    

    $$I_{av} = \frac{1}{T/2}\int_0^{T/2} i(t)\; dt =\frac{2}{T}\int_0^{T/2} \;I_... ...\pi t/T)\;dt =-\frac{2}{T}\frac{T}{2\pi} I_p\,\cos(2\pi t/T)\bigg\vert _0^{T/2}$$

    $$= \frac{I_p}{\pi}\left[\cos(0)-\cos(\pi)\right] =\frac{2}{\pi}\,I_p=0.637\,I_p$$ (50)

    
    

• Energy storage in capacitor $C$

  Given voltage $v(t)$ across and current $i(t)$ through a capacitor $C$, the associated energy is:

  $$w=\int_0^T p(t)dt=\int_0^T v(t)\; i(t) dt =\int_0^T v(t) \;C\frac{dv(t)}{dt} dt =C \int_0^V v\;dv=\frac{1}{2}CV^2$$ (51)

  where we have assumed $v(0)=0$ and $v(T)=V$.

  If $v(t)=V_p \sin(\omega t)$, then $i(t)=C\;dv(t)/dt=\omega C V_p\;\cos(\omega t)$, and the energy dissipated in period $T=2\pi/\omega$ is

  $$w= \int_0^T v(t)\; i(t) dt =V_p^2\omega C\int_0^T \sin(\omega t)\;\cos(\omega t)\,dt =\frac{V_p^2\omega C}{2} \int_0^T \sin(2\omega t) dt=0$$ (52)

  This results indicates that there is no energy dissipated over the complete period $T$. In the first and third quarters of the period $T$, the energy is stored in the electric field of the capacitor (equivalent to a battery being charged), but in the 2nd and 4th quarters of the period $T$, the energy is released from the capacitor to the rest of the circuit (equivalent to a battery delivering power).

• Energy storage in inductor $L$

  Given voltage $v(t)$ across and current $i(t)$ through an inductor $L$, the associated energy is

  $$w=\int_0^T p(t)dt=\int_0^T i(t)\; v(t) dt =\int_0^T i(t) \;L \frac{di(t)}{dt} dt =L \int_0^I i\;di=\frac{1}{2}LI^2$$ (53)

  where we have assumed $i(0)=0$ and $i(T)=I$.

  If $i(t)=I_p \sin(\omega t)$, then $v(t)= L\;di(t)/dt=L I_p\omega\;\cos(\omega t)$, and the energy dissipated in time period $T=2\pi/\omega$ is

  $$w=\int_0^T v(t)\;i(t) dt=I_p^2\omega L\int_0^T \sin(\omega t)\;\cos(\omega t)\,dt =\frac{I_p^2\omega L}{2} \int_0^T \sin(2\omega t) dt=0$$ (54)

  Again, no energy is dissipated by the inductor during the complete period $T$ of a sinusoidal voltage. In the first and third quarter of the period $T$, the energy is stored in the magnetic field of the inductor, but in the 2nd and 4th quarter of the period $T$, the energy is released from the inductor to the rest of the circuit.

The figure below shows the plots of the voltage across and current through the capacitor and inductor. We note that the voltage and current are $\pi/2$ or $90^\circ$ out of phase. Specifically,

• Capacitor: the voltage $v_C(t)=\sin(\omega t)$ (red) lags the current $i_C=\omega C\cos(\omega t)$ (green) by $\pi/2$ (or $90^\circ$).
• Inductor: the voltage $v_L(t)=\omega L \cos(\omega t)$ (green) leads the current $i_L=\sin(\omega t)$ (red) by $\pi/2$ (or $90^\circ$).

The figure below illustrates the energy flow in a circuit involving capacitor and inductor, as energy storing components:

Comparison of Energy storage in mechanical and electromagnetic systems:

• Electromagnetic energy:
  • Energy stored in a capacitor with capacity $C$ and voltage $V$:

    $$w=\frac{1}{2}CV^2$$ (55)

    
    

    $$[Farad][volt]^2=\frac{[Ampere][second]}{[Volt]}[Volt]^2$$

    $$= [Ampere][Volt][second]=[Watt][second]=[Joule]$$ (56)

    
    
  • Energy stored in an inductor with inductance $L$ and current $I$:

    $$w=\frac{1}{2}LI^2$$ (57)

    
    

    $$[Henry][Ampere]^2=\frac{[Volt][second]}{[Ampere]}[Ampere]^2$$

    $$= [Volt][Ampere][second]=[Watt][second]=[Joule]$$ (58)

    
    

• Mechanical energy:
  • Kinetic Energy: Energy stored in a mass of 1 kilogram moving with a velocity $u$ of 1 meter per second possesses 1/2 Joule of kinetic energy.

    $$w=\frac{1}{2}mu^2,\;\;\;\;\;\frac{[kilogram][meter]^2}{[second]^2}=[Joule]$$ (59)

    Another unit for energy is calorie: $1 \;$   cal$=4.2\;$Joules$$

  • Potential energy: Energy stored in a spring ($x=Cf=f/K$) of stiffness $K$ or compliance $C=1/K$ is

    $$w=\frac{1}{2}Cf^2=\frac{1}{2}K x^2,\;\;\;\;\; \frac{[meter]}{[Newton]}[Newton]^2=[meter][Newton]=[Joule]$$ (60)