ee1:aufgabe_7.2.6_mit_rechnung [MEXLE Wiki]

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<panel type="info" title="Exercise 5.2.6: Charging and Discharging of RC elements (exam task, ca. 11 % of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

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The circuit shown right is given with the following data:

  * $U = 10 ~{\rm V}$
  * $I = 4 ~{\rm mA}$
  * $R_1 = 100 ~\Omega, R_2 = 80 ~\Omega, R_3 = 50 ~\Omega, R_4 = 10 ~\Omega$
  * $C = 40 ~{\rm nF}$

At first, the voltage drop on the capacitor $u_C = 0$, and all switches are open. The switch S1 will be closed at $t = 0$.

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1. Determine the time constant $\tau$ for this charging process.

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  * What equivalent circuit can be found for the mentioned states of the switches?
  * What parameter do you need to determine $\tau$?
  * The charging current is flown through which component?

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<button size="xs" type="link" collapse="Loesung_7_2_6_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_1_Lösungsweg" collapsed="true">

The electrical components $R_1$, $R_2$, and $C$ are connected in series with a source $U$. 
The time constant $\tau$ is therefore: 
\begin{align*} 
\tau &= (R_1 + R_2) \cdot C \\ 
\tau &= 180 ~\Omega \cdot 40 ~{\rm nF} 
\end{align*}

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<button size="xs" type="link" collapse="Loesung_7_2_6_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_1_Endergebnis" collapsed="true"> 
\begin{align*} \tau = 7.2 ~{\rm µs}
\end{align*} \\ 
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2. What is the value of the voltage $u_C(t)$ drop over the capacitor $C$ at $t=10 ~{\rm µs}$?

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\begin{align*} 
U_C(t) = U           \cdot (1 - e^{-t/\tau}) \\ 
U_C(t) = 10 ~{\rm V} \cdot (1 - e^{-10 ~{\rm µs}/7.2 ~{\rm µs}}) 
\end{align*}

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<button size="xs" type="link" collapse="Loesung_7_2_6_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_2_Endergebnis" collapsed="true"> 

\begin{align*} U_C(t) = 7.506 ~{\rm V} \rightarrow 7.5 ~{\rm V} \end{align*} \\ </collapse>

3. What is the value of the energy, when the capacitor is fully charged?

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\begin{align*} 
W_C &= \frac{1}{2} C U^2 \\ 
    &= \frac{1}{2} \cdot 40~{\rm nF} \cdot (10~{\rm V})^2 
\end{align*}

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<button size="xs" type="link" collapse="Loesung_7_2_6_3_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_3_Endergebnis" collapsed="true"> 
\begin{align*} W_C = 2 ~{\rm µJ} \end{align*} \\ 
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4. Determine the new time constant when the switch $S_1$ will be opened and the switch $S_3$ will be closed simultaneously. 

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The capacitor $C$ discharges by the series connected resistors $R_2$ und $R_3$. 
\begin{align*} 
\tau &= (R_2 + R_3) \cdot C \\ 
     &= 130 ~\Omega \cdot 40 ~{\rm nF} 
\end{align*}

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<button size="xs" type="link" collapse="Loesung_7_2_6_4_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_4_Endergebnis" collapsed="true"> 
\begin{align*} \tau = 5.2 ~{\rm µs}
\end{align*} \\ </collapse>

5. When the capacitor is empty all switches will be opened. The switch $S_4$ will be closed at $t= 0$. \\ What is the voltage $u_C$ at the capacitor C after $t = 1 ~ {\rm µs}$?

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  * Through the current source there is a continuous flow of electric charge into the capacitor.
  * The resistors passed by the current on the way to the capacitor are irrelevant. They only increase the voltage of an ideal current source to guarantee the current.

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The voltage $U_C$ is in general: $U_C = \frac{Q}{C}$. In this case, the constant current I results in $Q = \int I {\rm d}t = I \cdot t$ 
\begin{align*} 
U_C(t)   &= \frac{Q}{C} \\ 
U_C(t)   &= \frac{I \cdot t}{C} \\ 
U_C(1μs) &= \frac{4~{\rm mA} \cdot 1~{\rm µs}}{40~{\rm nF}} 
          = \frac{4          \cdot 10^{-3}~{\rm A} \cdot 1\cdot 10^{-6}~{\rm s}}{40\cdot 10^{-9}~{\rm F}} \\ 
\end{align*}

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<button size="xs" type="link" collapse="Loesung_7_2_6_5_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_5_Endergebnis" collapsed="true"> 
\begin{align*} 
U_C(1~{\rm µs}) &= 1~{\rm V} \\ 
\end{align*} \\ 
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