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lab_electrical_engineering:2_capacitors:capacitors [2026/03/21 23:24] – mexleadminlab_electrical_engineering:2_capacitors:capacitors [2026/03/21 23:31] (current) – mexleadmin
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 {{drawio>lab_electrical_engineering:2_capacitors:Table-6-E6-series-Cap_V2.svg}} {{drawio>lab_electrical_engineering:2_capacitors:Table-6-E6-series-Cap_V2.svg}}
 <tabcaption Table-6-E6-series-Cap_V2 | E6 series for capacitors> </tabcaption> <tabcaption Table-6-E6-series-Cap_V2 | E6 series for capacitors> </tabcaption>
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 === RC network === === RC network ===
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 The capacitance of a capacitor is defined as the quotient of charge by voltage: The capacitance of a capacitor is defined as the quotient of charge by voltage:
 $$ C=\frac{Q}{U} $$ $$ C=\frac{Q}{U} $$
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 Because of the exponential function, charging is theoretically only complete after an infinitely long time. The capacitor voltage equals ${\rm 63}~{\rm\%~} U$ after ${\rm 1}\cdot \tau$, ${\rm 86}~{\rm\%~} U$ after ${\rm 2}\cdot \tau$, ${\rm 95}~{\rm\%~} U$ after ${\rm 3}\cdot \tau$, ${\rm 98}~{\rm\%~} U$ after ${\rm 4}\cdot \tau$, and ${\rm 99}~{\rm\%~} U$ after ${\rm 5}\cdot \tau$. It is assumed that the capacitor is fully charged after a time span $T = {\rm 5}\cdot \tau$ and the voltage across the capacitor has reached $U$. If the charged capacitor $C$ is discharged through a resistor $R$, the solution of the differential equation for the voltage is: \\ Because of the exponential function, charging is theoretically only complete after an infinitely long time. The capacitor voltage equals ${\rm 63}~{\rm\%~} U$ after ${\rm 1}\cdot \tau$, ${\rm 86}~{\rm\%~} U$ after ${\rm 2}\cdot \tau$, ${\rm 95}~{\rm\%~} U$ after ${\rm 3}\cdot \tau$, ${\rm 98}~{\rm\%~} U$ after ${\rm 4}\cdot \tau$, and ${\rm 99}~{\rm\%~} U$ after ${\rm 5}\cdot \tau$. It is assumed that the capacitor is fully charged after a time span $T = {\rm 5}\cdot \tau$ and the voltage across the capacitor has reached $U$. If the charged capacitor $C$ is discharged through a resistor $R$, the solution of the differential equation for the voltage is: \\
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 $$ u_{\rm C}({\rm t})=U\cdot e^{-\frac{t}{\tau}} $$ $$ u_{\rm C}({\rm t})=U\cdot e^{-\frac{t}{\tau}} $$
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 For the current accordingly: \\ For the current accordingly: \\
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 $$ i_{\rm C}({\rm t})=- \frac{U}{R}\cdot e^{-\frac{t}{\tau}} $$  $$ i_{\rm C}({\rm t})=- \frac{U}{R}\cdot e^{-\frac{t}{\tau}} $$ 
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 Now build the following circuit. Connect the function generator and the oscilloscope to the circuit as shown in <imgref Fig-10_V2-Osci-Function-gen>.  Now build the following circuit. Connect the function generator and the oscilloscope to the circuit as shown in <imgref Fig-10_V2-Osci-Function-gen>. 
  
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 {{drawio>lab_electrical_engineering:2_capacitors:Table-4-time-constant-meas_V2.svg}} {{drawio>lab_electrical_engineering:2_capacitors:Table-4-time-constant-meas_V2.svg}}
-<tabcaption Table-4-time-constant-meas_V2 | Capacitor meas. + time constant $\tau$> </tabcaption> \\ \\+<tabcaption Table-4-time-constant-meas_V2 | Capacitor meas. + time constant $\tau$> </tabcaption> \\
  
 Set the voltage $u_{\rm F}$ generated by the function generator to a unipolar square with amplitude 5 $V$ (i.e., no negative signal voltages occur!). The frequency on the function generator must be chosen so that the capacitor just fully charges and then fully discharges again. \\ Set the voltage $u_{\rm F}$ generated by the function generator to a unipolar square with amplitude 5 $V$ (i.e., no negative signal voltages occur!). The frequency on the function generator must be chosen so that the capacitor just fully charges and then fully discharges again. \\
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 $f_{\rm 1} =~{\rm ...............}$ $f_{\rm 1} =~{\rm ...............}$
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 Sketch the voltages measured with the oscilloscope for $u_{\rm F}$, $u_{\rm C}$, and $u_{\rm R}$ in the following screen diagram. Also enter alongside the screen drawings the set $\frac{V}{\rm DIV}$ of the channels and the $\frac{T}{\rm DIV}$ of the time base. \\ Sketch the voltages measured with the oscilloscope for $u_{\rm F}$, $u_{\rm C}$, and $u_{\rm R}$ in the following screen diagram. Also enter alongside the screen drawings the set $\frac{V}{\rm DIV}$ of the channels and the $\frac{T}{\rm DIV}$ of the time base. \\
  
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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-Draw **tangents** in the screen diagram for the start of charging and the start of discharging. What is the charging current or discharging current at the beginning?\\+\\
  
 +Draw **tangents** in the screen diagram for the start of charging and the start of discharging. \\ 
 +What is the charging current or discharging current at the beginning?\\
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 +\\ \\ ${\rm ........................................................................................}$
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 \\ \\ ${\rm ........................................................................................}$  \\ \\ ${\rm ........................................................................................}$ 
 \\ \\ ${\rm ........................................................................................}$  \\ \\ ${\rm ........................................................................................}$ 
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