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Excercise 7.2.6: temperature dependent resistor of a winding (exam task, ca. 11% of a 60 minute exam, WS2020)
The adjacent circuit with the following data is given:
- $U = 10 V$
- $I = 4 mA$
- $R_1 = 100 \Omega, R_2 = 80 \Omega, R_3 = 50 \Omega, R_4 = 10 \Omega$
- $C = 40 nF$
At first the capacitor is empty and all switches are open. The switsch S1 will be closed at t=0.
1. Determine the time constant $\tau$ for this charging process.
- 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?
The electrical components $R_1$, $R_2$ und $C$ 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 nF \end{align*}
\begin{align*} \tau = 7,2 µs \end{align*}
2. How high is the voltage at the capacitor $C$ when $t=10 µs$?
\begin{align*} U_C(t) = U \cdot (1 - e^{-t/\tau}) \\ U_C(t) = 10 V \cdot (1 - e^{-10 µs/7,2 µs}) \end{align*}
\begin{align*} U_C(t) = 7,506 V -> 7,5 V \end{align*}
3. How high is the energy when the capacitor is fully charged?
\begin{align*} W_C &= \frac{1}{2}CU^2 \\ &= \frac{1}{2} \cdot 40nF \cdot (10V)^2 \end{align*}
\begin{align*} W_C = 2 µJ \end{align*}
4. Determine the new time constant when the capacitor is fully charged. The switch S1 will be opened whereas the switch S2 will be closed simultaneously.
The capacitor $C$ discharges by the series connected resistors $R_2$ und $R_3$. \begin{align*} \tau &= (R_2 + R_3) \cdot C \\ \tau &= 130 \Omega \cdot 40 nF \end{align*}
\begin{align*} \tau = 5,2 µs \end{align*}
5. When the capacitor is empty all switches will be opend. The switch S4 will be closed at $t = 1μs$.
How high is the voltage at the capacitor C?
- Through the current source there is a continuous flow of elctric 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.
The voltage $U_C$ is in general: $U_C = \frac{Q}{C}$. In this case the constant current I results in $Q = \int I dt = I \cdot t$ \begin{align*} U_C(t) &= \frac{Q}{C} \\ U_C(t) &= \frac{I \cdot t}{C} \\ U_C(1μs) &= \frac{4mA \cdot 1μs}{40nF} = \frac{4 \cdot 10^{-3}A \cdot 1\cdot 10^{-6}s}{40\cdot 10^{-9}F} \\ \end{align*}
\begin{align*} U_C(1μs) &= 1V \\ \end{align*}