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electrical_engineering_and_electronics_2:block06 [2026/04/14 01:01] – mexleadminelectrical_engineering_and_electronics_2:block06 [2026/04/21 03:48] (current) – mexleadmin
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   - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations.   - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations.
  
-==== Ideal Inductivity L ====+==== Ideal Capacity C ====
  
-A similar approach is done for the ideal inductivity.  +Also here, we start with the basic definition of the instantaneous voltage
-We again start with the basic definition of the instantaneous voltage+
  
-\begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}+\begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}
  
-With the defining formula for inductivity, we get: +With the defining formula for the capacity, we get: 
 \begin{align*}  \begin{align*} 
-           \color{blue}{u_{\rm L}(t)} &=  L\cdot    {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\rm d}t}} \\  +\color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\  
-\rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int  \color{blue}{u_{\rm L}(t)}       {\rm d}t \\  +                    &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) 
-                                &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) +
 \end{align*} \end{align*}
  
-This leads to an instantaneous power $p_L(t)$ of+This leads to an instantaneous power $p_C(t)$ of
  
 \begin{align*}  \begin{align*} 
-p_L(t) &=   \color{blue}{u(t)}            \cdot \color{red}{i(t)} \\  +p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\  
-       &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin(          \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  +       &= 2\cdot U^2 \omega C  \cdot \sin(         \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  
-       &= - {{U^2}\over{\omega L}}        \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ +       &= +      U^2 \omega C  \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ 
 \end{align*} \end{align*}
- 
-Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}}  "$\sin(2x) = 2 \sin(x)\cos(x)$") was used. 
  
 Also, this result is interesting: Also, this result is interesting:
  
   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
-  - Therefore, the average value of $p_L(t)=0$+  - Therefore, also the average value of $p_C(t)=0$
  
-==== Ideal Capacity C ====+==== Ideal Inductivity L ====
  
-Also here, we start with the basic definition of the instantaneous voltage+A similar approach is done for the ideal inductivity.  
 +We again start with the basic definition of the instantaneous voltage
  
-\begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}+\begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}
  
-With the defining formula for the capacity, we get: +With the defining formula for inductivity, we get: 
 \begin{align*}  \begin{align*} 
-\color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\  +           \color{blue}{u_{\rm L}(t)} &=  L\cdot    {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\rm d}t}} \\  
-                    &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) +\rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int  \color{blue}{u_{\rm L}(t)}       {\rm d}t \\  
 +                                &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) 
 \end{align*} \end{align*}
  
-This leads to an instantaneous power $p_C(t)$ of+Since we assume pure AC signals the integration constant has to be 0. 
 +\\ 
 +This formulas lead to an instantaneous power $p_L(t)$ of
  
 \begin{align*}  \begin{align*} 
-p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\  +p_L(t) &=   \color{blue}{u(t)}            \cdot \color{red}{i(t)} \\  
-       &= 2\cdot U^2 \omega C  \cdot \sin(         \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  +       &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin(          \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  
-       &= +      U^2 \omega C  \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ +       &= - {{U^2}\over{\omega L}}        \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ 
 \end{align*} \end{align*}
 +
 +Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}}  "$\sin(2x) = 2 \sin(x)\cos(x)$") was used.
  
 Again this result leads to: Again this result leads to:
  
   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
-  - Therefore, also the average value of $p_C(t)=0$+  - Therefore, the average value of $p_L(t)=0$
  
   * Instantaneous values of power at $R$, $L$, $C$   * Instantaneous values of power at $R$, $L$, $C$
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 Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller. Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller.
  
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 Another explanation of the power factor can be seen here: Another explanation of the power factor can be seen here: