Energy Dissipation/Storage in Linear Elements

By definition, $v=dw/dq$, $i=dq/dt$, therefore the electric power associated with an element is

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

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: When a DC voltage $V$ is applied across $R$, the current through it is $I=V/R$, the power consumption is $p=VI$ and the energy dissipation during time period $0 \sim T$ is
  
  $$w=pT=VIT=\frac{V^2}{R}T=I^2RT$$
  
  
  When an AC voltage $v(t)$ is applied across $R$, the current through it is $i(t)=v(t)/R$, power consumption is $p(t)=v(t)i(t)$ and the energy dissipated during time period $0 \sim T$ is
  
  $$w=\int_0^T p(t) dt=\int_0^T v(t)\; i(t) dt =\frac{1}{R}\;\int_0^T v^2(t)\; dt =R\;\int_0^T i^2(t)\; dt$$
  
  
  This energy is converted irreversibly from electrical energy to heat which cannot be converted back to electrical energy. The rate of dissipation is
  
  $$p(t)=\frac{dw}{dt}=\frac{v^2(t)}{R}=i^2(t)R$$
  
  

  When $v(t)=V_p \sin(2\pi f t)=V_p \sin(\omega t)$ and $i(t)=v(t)/R=V_p\sin(2\pi ft)/R=V_p\sin(\omega t)/R$ (where $V_p$ is the peak value of the voltage, $f$ is the frequency and $\omega=2\pi f$ is the angular frequency), the energy dissipated in time period $T=1/f=2\pi/\omega$ is
  

  $$w=\frac{1}{R}\int_0^T v^2(t) dt=\frac{V_p^2}{R}\int_0^T sin^2... ...1-cos (2\omega t)] dt=\frac{V_p^2T}{2R} =\frac{V_{rms}^2T}{R}$$
  
  
  where
  
  $$V_{rms}=\sqrt{\frac{1}{T}\int_0^T v^2(t) dt }=\frac{V_p}{\sqrt{2}}=0.707 V_p$$
  
  
  is the effective or root-mean-square (RMS) voltage of $v(t)$, representing the required DC voltage across the resistor for it to dissipate the same amount of energy.

  

  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:
    
    $$w=\frac{1}{2}C V^2$$
    
    
    When $v(t)=V_p \sin(\omega t)$, $i(t)=C\;dv(t)/dt=C V_p\omega\;cos(\omega t)$, the energy dissipated in period $T=2\pi/\omega$ is
    
    $$w= \int_0^T v(t) i(t) dt=V_p^2\omega C\int_0^T sin(\omega t) ... ...a t) dt =\frac{V_p^2\omega C}{2} \int_0^T sin(2\omega t) dt=0$$
    
    
    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:
    
    $$w=\frac{1}{2}L I^2$$
    
    

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

    $$w= \int_0^T v(t) i(t) dt=I_p^2\omega L\int_0^T sin(\omega t) ... ...a t) dt =\frac{I_p^2\omega L}{2} \int_0^T sin(2\omega t) dt=0$$
    
    
    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).