Energy Dissipation/Storage in Linear Elements

The electric power associated with an element is:

$$p(t)=v(t) i(t)$$

where $i(t)$ is the current going through it and $v(t)$ the voltage across it. Energy is power integrated over time:

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

• Energy Dissipation: For a resistor with $v(t)=R\;i(t)$ and $i(t)=v(t)/R$, the energy dissipated in time period $(0 \sim T)$ is:
  
  $$w=\int_0^T v(t)\; i(t) dt=\frac{1}{R}\;\int_0^T v^2(t)\; dt =R\;\int_0^T i^2(t)\; dt$$
  
  
  This energy is converted irreversibly from electrical energy to heat which cannot be converted back to electrical energy. The rate of dissipation is:
  
  $$p(t)=\frac{d}{dt} w=\frac{v^2(t)}{R}=i^2(t)R$$
  
  

  When the current and voltage are both constant $i(t)=I$ and $v(t)=V$, we have $w=V^2T/R=I^2RT$.

  Example: When $v(t)=V_m \sin(\omega t)$ and $i(t)=v(t)/R=V_m\sin(\omega t)/R$, the energy dissipated in time period $T=2\pi/\omega$ is
  

  $$w=\frac{1}{R}\int_0^T v^2(t) dt=\frac{V_m^2}{R}\int_0^T sin^2... ...ac{V_m^2}{2R}\int_0^T (1-cos (2\omega t)) dt=\frac{V_m^2T}{2R}$$
  
  

  

  Some useful trigonometric identities:
  

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

• Energy Storage:
  • Capacitor
    
    $$w=\int_{t_0}^{t_1} v(t)\; i(t) dt=\int_{t_0}^{t_1} v(t) C \fr... ...(t)}{dt} dt =C \int_{v_1}^{v_2} v dv=\frac{C}{2}(v_1^2-v_0^2)$$
    
    
    Here we have assumed $v(t_0)=v_0$ and $v(t_1)=v_1$. We see that capacitance $C$ is a measure of the capacitor's ability to store energy in electric field (separated charge in terms of voltage $v$). In particular, under zero initial condition $v_0=v(t_0)=0$ and $v_1=v(t_1)=V$, the potential energy stored in the capacitor is proportional to $C$ and voltage $V$ squared:
    
    $$w=\frac{1}{2}C V^2$$
    
    

    Example: When $v(t)=V_m \sin(\omega t)$, $i(t)=C\;dv(t)/dt=C V_m \omega\;cos(\omega t)$, the energy dissipated in period $T=2\pi/\omega$ is
    

    $$w=\int_0^T v(t) i(t) dt=V_m^2 \omega C\int_0^T\sin(\omega t)\... ... t)dt =\frac{V_m^2 \omega C}{2} \int_0^T\sin(2\omega t) dt=0$$
    
    
    This results indicates that there is no energy dissipated over the complete period $T$, as in the first and third $T/4$ the energy is stored in the capacitor (equivalent to a battery being charged), but in the 2nd and 4th $T/4$ the energy is released from the capacitor again (equivalent to a battery delivering power).

    

  • Inductor
    
    $$w=\int_{t_0}^{t_1} v(t)\; i(t) dt=\int_{t_0}^{t_1} i(t) L \fr... ...(t)}{dt} dt =L \int_{i_1}^{i_2} i di=\frac{L}{2}(i_1^2-i_0^2)$$
    
    
    Here we have assumed $i(t_0)=i_0$ and $i(t_1)=i_1$. We see that inductance $L$ is a measure of the inductor's ability to store energy in magnetic field (moving charge in terms of current $i$). In particular, under zero initial condition $i_0=i(t_0)=0$ and $i_1=i(t_1)=I$, the potential energy stored in the inductor is proportional to $L$ and the current $I$ squared:
    
    $$w=\frac{1}{2}L I^2$$
    
    

    Example: When $i(t)=I_m sin(\omega t)$, $v(t)= L\;di(t)/dt=L I_m \omega\;cos(\omega t)$, the energy dissipated in time period $T=2\pi/\omega$ is
    

    $$w= \int_0^T v(t) i(t) dt=I_m^2\omega L\int_0^T sin(\omega t) ... ...a t) dt =\frac{I_m^2\omega L}{2} \int_0^T sin(2\omega t) dt=0$$
    
    
    Similarly, for a sinusoidal current of period $T$, no energy will be dissipated by the inductor during the complete period $T$. In both of the cases of capacitor and inductor, the energy is converted into potential energy stored in either the electric or magnetic field of the element, instead of being dissipated (converted to heat).
  

Comparison with mechanical systems:

The work of a mechanical system does is $w=\int_0^X f(x)\; dx$ where $f(x)$ is force and $X$ is displacement.

• Potential energy: A spring can be described by Hooke's law $f=Kx$ where $K$ is the stiffness, or $x=Cf$ where $C=1/K$ is the compliance. The potential energy stored in the spring is:
  
  $$w=\int_0^X f(x)\;dx=\int_0^F f\;C\;df=\frac{1}{2}CF^2$$
  
  
  i.e., the compliance $C$ is a measure of the spring's ability to store potential energy (the less stiff, the more potential energy can be stored in the spring with the same force).
• Kinetic energy: A mass moving at a velocity $v=dx/dt$ has kinetic energy
  
  $$w=\int_0^X f(x)\;dx=\int_0^T f(t)\;\frac{dx}{dt} dt=\int_0^T f(t)\;v\;dt$$
  
  
  But as $f=ma=m\;dv/dt$, the kinetic energy becomes:
  
  $$w=\int_0^T f(t)\;v\;dt=\int_0^T m \frac{dv}{dt}\;v\;dt=m \int_0^V v\;dv =\frac{1}{2}mV^2$$
  
  
  i.e., the mass $m$ of a body is a measure of the body's ability to store kinetic energy (the more mass, the more kinetic energy can be stored in the body with the same velocity).

Examples:

• A mechanical system composed of a spring $K$ and a mass $m$ can be described by:
  
  $$m\frac{d^2x}{dt^2}+Kx=0,\;\;\;\;\mbox{i.e.}\;\;\;\; \frac{d^2x}{dt^2}+\frac{K}{m}x=0$$
  
  
  where $x$ is the horizontal displacement of the mass. The homogeneous solution of the differential equation can be found to be
  
  $$x(t)=sin(\omega t),\;\;\;\;\dot{x}(t)=\omega\;cos(\omega t), ... ...omega^2\;sin(\omega t), \;\;\;\;\omega=\sqrt{K/m}=1/\sqrt{Cm}$$
  
  
  where $C=1/K$ is the compliance. When the displacement is zero $x(t)=sin(\omega t)=0$, the velocity $\dot{x}(t)=\omega\;cos(\omega t)=\pm \omega$ is maximal. This corresponds to the fact that when the potential energy stored in the spring is zero, the kinetic energy is maximal. On the other hand, when the displacement is maximal $x(t)=sin(\omega t)=\pm 1$, the velocity $\dot{x}(t)=\omega\;cos(\omega t)=0$ is zero. This corresponds to the fact that when the potential energy stored in the spring is maximal, the kinetic energy is zero.

• An electrical system composed of a capacitor $C$ and an inductor $L$ can be described by:
  
  $$i_C=C\frac{dv}{dt},\;\;\;v=L\frac{di_L}{dt}$$
  
  
  where $v$ is the voltage across both components, and $i_C$ and $i_L$ are currents through $C$ and $L$, respectively. As $i_C+i_L=0$, i.e. $i_C=-i_L$, we have
  
  $$v=L\frac{di_L}{dt}=L\frac{d}{dt}\;(-C\frac{dv}{dt}),\;\;\; \mbox{i.e.}\;\;\;\;\frac{d^2v}{dt^2}+\frac{1}{LC}v=0$$
  
  
  The homogeneous solution of the differential equation can be found to be
  
  $$v(t)=sin(\omega t),\;\;\;\;\dot{v}(t)=\omega\;cos(\omega t), ... ...ot{v}(t)=-\omega^2\;sin(\omega t), \;\;\;\;\omega=1/\sqrt{LC}$$
  
  
  When the energy stored in the capacitor is zero $v(t)=sin(\omega t)=0$, the energy stored in the inductor is maximal $i(t)=C\dot{v}(t)=\omega C\;cos(\omega t)=\pm \omega C$; but when the energy stored in the capacitor is maximal $v(t)=sin(\omega t)=\pm 1$, the energy stored in the inductor is zero $i(t)=C\dot{v}(t)=\omega C\;cos(\omega t)=0$.

In both cases, the energy in the system is converted back and forth between different forms (potential vs. kinetic, electrical vs. magnetic), while the total amount is always reserved.

However, when a dash-pot (causing friction proportional to speed $\dot{x}$) is added (in parallel to the spring) in the mechanical system, and a resistor is added in series with the electrical circuit, the energy is dissipated (converted to heat) in both systems:

$$\frac{d^2x}{dt^2}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$$

$$\frac{d^2v}{dt^2}+\frac{R}{L}\frac{dv}{dt}+\frac{1}{LC}v=0$$

The corresponding solution of the DEs will be decaying sinusoidal, indicating the dissipation of the energy in the system.