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electrical_engineering_1:dc_circuit_transients [2023/11/30 00:06] mexleadmin [Bearbeiten - Panel] |
electrical_engineering_1:dc_circuit_transients [2023/12/02 00:55] mexleadmin [Exercises] |
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Zeile 474: | Zeile 474: | ||
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- | # | + | # |
+ | |||
+ | The following circuit shows a charging/ | ||
+ | |||
+ | The values of the components shall be the following: | ||
+ | * $R_1 = 1 \rm k\Omega$ | ||
+ | * $R_2 = 2 \rm k\Omega$ | ||
+ | * $R_3 = 3 \rm k\Omega$ | ||
+ | * $C = 1 \rm \mu F$ | ||
+ | * $S_1$ and $S_2$ are opened in the beginning (open-circuit) | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | 1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\ | ||
+ | |||
+ | 1.1 What is the value of the time constant $\tau_1$? | ||
+ | |||
+ | # | ||
+ | |||
+ | The time constant $\tau$ is generally given as: $\tau= R\cdot C$. \\ | ||
+ | Now, we try to determine which $R$ and $C$ must be used here. \\ | ||
+ | To find this out, we have to look at the circuit when $S_1$ gets closed. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | We see that for the time constant, we need to use $R=R_1 + R_2$. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | \tau_1 &= R\cdot C \\ | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | 1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! | ||
+ | |||
+ | # | ||
+ | |||
+ | To get a general formula, we again take a look at the circuit, but this time with the voltage arrows. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$\\ | ||
+ | The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/ | ||
+ | Therefore, $u_{R2}$ can be written as: | ||
+ | |||
+ | \begin{align*} | ||
+ | u_{R2} &= R_2 \cdot i_{R2} \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. | ||
+ | At this point, the switch $S_2$ will be closed. \\ Calculate $t_1$! | ||
+ | |||
+ | # | ||
+ | |||
+ | We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/ | ||
+ | Therefore we get $t_1$ by: | ||
+ | |||
+ | \begin{align*} | ||
+ | u_C = 4/5 \cdot U_1 & | ||
+ | 4/5 & | ||
+ | e^{-t/ | ||
+ | | ||
+ | t &= -\tau \cdot \rm ln (1/5) \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | t & | ||
+ | \end{align*} | ||
+ | # | ||
+ | |||
+ | 3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5 ~\rm V$ and $U_2 = 10 ~\rm V$. | ||
+ | |||
+ | 3.1 What is the new time constant $\tau_2$? | ||
+ | |||
+ | # | ||
+ | |||
+ | Again the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ | ||
+ | Again, we try to determine which $R$ and $C$ must be used here. \\ | ||
+ | To find this out, we have to look at the circuit when both $S_1$ and $S_2$ are closed. \\ | ||
+ | In this case, we can " | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | We see that for the time constant, we now need to use $R=R_1 || R_3 + R_2$. | ||
+ | |||
+ | \begin{align*} | ||
+ | \tau_2 &= R\cdot C \\ | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | \tau_2 &= 2.75~\rm ms \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | 3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. | ||
+ | |||
+ | 3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. | ||
+ | # | ||
+ | |||
+ | {{page> | ||
+ | |||
+ | # | ||
Zeile 511: | Zeile 634: | ||
- | {{page> | ||