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electrical_engineering_1:dc_circuit_transients [2023/12/02 00:50] – [Exercises] mexleadminelectrical_engineering_1:dc_circuit_transients [2023/12/02 00:55] – [Exercises] mexleadmin
Zeile 483: Zeile 483:
   * $R_3 = 3 \rm k\Omega$   * $R_3 = 3 \rm k\Omega$
   * $C   = 1 \rm \mu F$   * $C   = 1 \rm \mu F$
-  * $S_1$ and $S_2$ are opened (open-circuit)+  * $S_1$ and $S_2$ are opened in the beginning (open-circuit)
  
 {{drawio>electrical_engineering_1:Exercise522setup.svg}} {{drawio>electrical_engineering_1:Exercise522setup.svg}}
Zeile 569: Zeile 569:
 Again the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ 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. \\ 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.+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 "fold" over the resistor $R_3$ to the left-side of the capacitor $C$.
  
 {{drawio>electrical_engineering_1:Exercise522sol4.svg}} {{drawio>electrical_engineering_1:Exercise522sol4.svg}}
Zeile 578: Zeile 579:
 \tau_2 &= R\cdot C \\ \tau_2 &= R\cdot C \\
        &= (R_1 || R_3 + R_2) \cdot C \\        &= (R_1 || R_3 + R_2) \cdot C \\
-       &= {{\rm  1 ~k\Omega \cdot 3 ~k\Omega }\over{\rm  1 ~k\Omega + 3 ~k\Omega}} + 2 ~{\rm k\Omega}) \cdot 1 ~{\rm \mu F} \\+       &= \left({{\rm  1 ~k\Omega \cdot 3 ~k\Omega }\over{\rm  1 ~k\Omega + 3 ~k\Omega}} + 2 ~{\rm k\Omega}\right) \cdot 1 ~{\rm \mu F} \\
 \end{align*} \end{align*}