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| | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor #@TaskText_HTML@# |
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| ====== Experiment 1 ====== | A machine-vision inspection system on a production line uses a short high-voltage flash pulse. For this purpose, an energy-storage capacitor is charged from a DC source and must be safely discharged before maintenance. |
| ===== DC circuit theory ===== | |
| ==== Linear and non-linear resistors ==== | |
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| ^ Name ^ <wrap onlyprint> \\ \\ </wrap> ^ | Data: \begin{align*} C &= 1~{\rm \mu F} \\ W_e &= 0.1~{\rm J} \\ I_{\rm max} &= 100~{\rm mA} \\ R_i &= 10~{\rm M\Omega} \end{align*} |
| ^ Student ID number ^ <wrap onlyprint> \\ \\ </wrap> ^ | |
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| === Equipment used === | 1. What voltage must the capacitor have so that it stores the required energy? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~101~@# <WRAP leftalign> \begin{align*} W_e &= \frac{1}{2} C U^2 \\ U &= \sqrt{\frac{2W_e}{C}} = \sqrt{\frac{2 \cdot 0.1~{\rm J}}{1 \cdot 10^{-6}~{\rm F}}} \\ &= \sqrt{200000}~{\rm V} \approx 447.2~{\rm V} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~101~@# \begin{align*} U = 447.2~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| * Bench power supply GPS 3303 | 2. The charging current must not exceed $100~\rm mA$ at the start of charging. What charging resistor is required? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~102~@# <WRAP leftalign> At the beginning of charging, the capacitor behaves like a short circuit, so \begin{align*} i_{C{\rm max}} = i_C(t=0) = \frac{U}{R} \end{align*} Thus, \begin{align*} R &\ge \frac{U}{I_{\rm max}} = \frac{447.2~{\rm V}}{0.1~{\rm A}} \\ &\approx 4472~{\rm \Omega} = 4.47~{\rm k\Omega} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~102~@# \begin{align*} R \ge 4.47~{\rm k\Omega} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| * Digital multimeter Agilent U1241A | |
| * Breadboard GL-36 | |
| * Decade resistance box RD-1000, $\pm 1 \%$ | |
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| The aim of this experiment is to become familiar with and investigate the following: | 3. How long does the charging process take until the capacitor is practically fully charged? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~103~@# <WRAP leftalign> The time constant is \begin{align*} T = RC = 4.47~{\rm k\Omega} \cdot 1~{\rm \mu F} = 4.47~{\rm ms} \end{align*} In engineering practice, a capacitor is considered practically fully charged after about $5T$: \begin{align*} t \approx 5T = 5 \cdot 4.47~{\rm ms} = 22.35~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~103~@# \begin{align*} t \approx 22.35~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| * assembling simple circuits on the GL-36 breadboard | 4. Give the time-dependent capacitor voltage and the voltage across the charging resistor. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~104~@# <WRAP leftalign> For the charging process: \begin{align*} u_C(t) &= U\left(1-e^{-t/T}\right) \\ u_R(t) &= Ue^{-t/T} \end{align*} with \begin{align*} U &= 447.2~{\rm V} \\ T &= 4.47~{\rm ms} \end{align*} So the capacitor voltage rises exponentially from $0$ to $447.2~\rm V$, while the resistor voltage falls exponentially from $447.2~\rm V$ to $0$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~104~@# \begin{align*} u_C(t) &= 447.2\left(1-e^{-t/4.47{\rm ms}}\right)~{\rm V} \\ u_R(t) &= 447.2\,e^{-t/4.47{\rm ms}}~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| * carrying out measurements with the Agilent U1241A digital multimeter | |
| * using resistor standard series and the associated colour codes | |
| * measuring resistances, voltages and currents | |
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| ==== General measurement techniques ==== | 5. After charging, the capacitor is disconnected from the source. Its leakage can be modeled by an internal resistance of $10~\rm M\Omega$. After what time has the stored energy dropped to one half, and what is the capacitor voltage then? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~105~@# <WRAP leftalign> Half the energy means \begin{align*} W_e' = 0.5W_e \end{align*} Since \begin{align*} W_e = \frac{1}{2}CU^2 \end{align*} the voltage at half energy is \begin{align*} U' = \frac{U}{\sqrt{2}} = \frac{447.2~{\rm V}}{\sqrt{2}} = 316.2~{\rm V} \end{align*} For the discharge through the internal resistance: \begin{align*} u_C(t) = Ue^{-t/T_2} \end{align*} with \begin{align*} T_2 = R_iC = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s} \end{align*} Set $u_C(t)=U'$: \begin{align*} Ue^{-t/T_2} &= U' \\ t &= T_2 \ln\left(\frac{U}{U'}\right) \\ &= 10~{\rm s}\cdot\ln\left(\frac{447.2}{316.2}\right) \\ &\approx 3.47~{\rm s} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~105~@# \begin{align*} U' = 316.2~{\rm V} \\ t = 3.47~{\rm s} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| === Voltage measurement === | 6. The fully charged capacitor is discharged through the charging resistor before maintenance. How long does the discharge take, and how much energy is converted into heat in the resistor? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~106~@# <WRAP leftalign> The discharge time constant through the same resistor is again \begin{align*} T = RC = 4.47~{\rm ms} \end{align*} Thus the practical discharge time is \begin{align*} t \approx 5T = 22.35~{\rm ms} \end{align*} The complete stored capacitor energy is converted into heat in the resistor: \begin{align*} W_R = W_e = 0.1~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~106~@# \begin{align*} t \approx 22.35~{\rm ms} \\ W_R = 0.1~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| Procedure for voltage measurement: | #@TaskEnd_HTML@# |
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| - Set the meter to the largest voltage range (check whether direct voltage or alternating voltage is to be measured; not necessary in auto range). | {{tag>rc_circuit thevenin_equivalent transient_response sensor_interface industrial_electronics chapter1_1}}{{include_n>300}} |
| - Connect the test leads to the correct meter sockets (the sockets marked COM and V). | |
| - Connect the test leads to the component under test with the correct polarity, so that the meter is connected in parallel with the component. | |
| - Read the measured value. | |
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| === Current measurement === | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Sensor Input Buffer: Source, T-Network and Capacitor #@TaskText_HTML@# |
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| Procedure for current measurement: | A 12 V industrial sensor electronics unit feeds a buffered measurement node through a resistor T-network. A capacitor smooths the node voltage. At first, the load is disconnected. After the capacitor is fully charged, a measurement load is connected by a switch. |
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| - Set the meter to the largest current range (check whether direct current or alternating current is to be measured; not necessary in auto range). | Data: \begin{align*} U &= 12~{\rm V} \\ R_1 &= 2~{\rm k\Omega} \\ R_2 &= 10~{\rm k\Omega} \\ R_3 &= 3.33~{\rm k\Omega} \\ C &= 2~{\rm \mu F} \\ R_L &= 5~{\rm k\Omega} \end{align*} |
| - Connect the test leads to the correct meter sockets (the sockets marked COM and $\mu{\rm A}.{\rm mA}$). | |
| - Connect the test leads to the component under test with the correct polarity, so that the meter is connected in series with the component. | |
| - Read the measured value. | |
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| === Resistance measurement === | Initially, the capacitor is uncharged and the switch is open. |
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| Procedure for resistance measurement: | 1. What is the capacitor voltage after it is fully charged? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~201~@# <WRAP leftalign> Using the equivalent voltage source of the network on the left-hand side, the open-circuit voltage is \begin{align*} U_{0e} &= \frac{R_2}{R_1+R_2}\,U \\ &= \frac{10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}}\cdot 12~{\rm V} \\ &= 10~{\rm V} \end{align*} After full charging, the capacitor voltage equals this voltage. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~201~@# \begin{align*} U_C = U_{0e} = 10~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| - Set the meter to resistance measurement. | 2. How long does the charging process take? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~202~@# <WRAP leftalign> The internal resistance seen by the capacitor is \begin{align*} R_{ie} &= R_3 + (R_1 \parallel R_2) \\ &= 3.33~{\rm k\Omega} + \frac{2~{\rm k\Omega}\cdot 10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}} \\ &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\ &= 5.00~{\rm k\Omega} \end{align*} So the time constant is \begin{align*} T &= R_{ie}C = 5.00~{\rm k\Omega}\cdot 2~{\rm \mu F} = 10~{\rm ms} \end{align*} Practical charging time: \begin{align*} t \approx 5T = 50~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~202~@# \begin{align*} R_{ie} = 5.00~{\rm k\Omega} \\ t \approx 50~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| - Connect the resistor to be measured to the corresponding sockets on the meter (the sockets marked COM and $\Omega$). | |
| - Read the measured value. | |
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| === Digital multimeter Agilent U1241A === | 3. Give the time-dependent capacitor voltage. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~203~@# <WRAP leftalign> The charging law is \begin{align*} u_C(t) &= U_{0e}\left(1-e^{-t/T}\right) \\ &= 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} \end{align*} So the capacitor voltage rises exponentially from $0~\rm V$ to $10~\rm V$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~203~@# \begin{align*} u_C(t) = 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| The Agilent U1241A multimeter has automatic range selection. The following measuring ranges are available: | 4. After the capacitor is fully charged, the switch is closed and the load resistor is connected. What is the stationary load voltage? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~204~@# <WRAP leftalign> Now use a second equivalent voltage-source step. The Thevenin source seen by the load has \begin{align*} U_{0e} &= 10~{\rm V} \\ R_{ie} &= 5.00~{\rm k\Omega} \end{align*} Thus, the stationary load voltage is \begin{align*} U_C' = U_{0e}' &= \frac{R_L}{R_{ie}+R_L}\,U_{0e} \\ &= \frac{5~{\rm k\Omega}}{5~{\rm k\Omega}+5~{\rm k\Omega}}\cdot 10~{\rm V} \\ &= 5~{\rm V} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~204~@# \begin{align*} U_L = 5~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| ^ Function ^ Range ^ Accuracy ^ | 5. How long does it take until this new stationary state is practically reached? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~205~@# <WRAP leftalign> The new internal resistance is \begin{align*} R_{ie}' &= R_{ie}\parallel R_L \\ &= 5.00~{\rm k\Omega}\parallel 5.00~{\rm k\Omega} \\ &= 2.50~{\rm k\Omega} \end{align*} Hence the new time constant is \begin{align*} T' &= R_{ie}'C = 2.50~{\rm k\Omega}\cdot 2~{\rm \mu F} = 5~{\rm ms} \end{align*} Practical settling time: \begin{align*} t \approx 5T' = 25~{\rm ms} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~205~@# \begin{align*} R_{ie}' = 2.50~{\rm k\Omega} \\ t \approx 25~{\rm ms} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| | DC voltage | $0 \ldots 1000 ~{\rm V}$ | $\pm 0.1 \%$ | | |
| | AC voltage | $0 \ldots 1000 ~{\rm V}$ | $\pm 1 \%$ | | |
| | DC current | $0 \ldots 10 ~{\rm A}$ | $\pm 0.2 \%$ | | |
| | AC current | $0 \ldots 10 ~{\rm A}$ | $\pm 1 \%$ | | |
| | Resistance | $0 \ldots 100 ~{\rm M}\Omega$ | $\pm 0.3 \%$ | | |
| | Capacitance | $0 \ldots 10 ~{\rm mF}$ | $\pm 1.2 \%$ | | |
| | Frequency | $30 ~{\rm Hz} \ldots 100 ~{\rm kHz}$ | $\pm 0.3 \%$ | | |
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| === Physical quantities and units used === | 6. Give the time-dependent load voltage after the switch is closed. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~206~@# <WRAP leftalign> At the switching instant, the capacitor voltage cannot jump. Therefore: \begin{align*} u_L(0^+) &= 10~{\rm V} \\ u_L(\infty) &= 5~{\rm V} \end{align*} The voltage therefore decays exponentially toward the new final value: \begin{align*} u_L(t) &= u_L(\infty) + \left(u_L(0^+)-u_L(\infty)\right)e^{-t/T'} \\ &= 5 + 5e^{-t/5{\rm ms}}~{\rm V} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~206~@# \begin{align*} u_L(t) = 5 + 5e^{-t/5{\rm ms}}~{\rm V} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| ^ Quantity ^ Symbol ^ Unit ^ Unit symbol ^ | #@TaskEnd_HTML@# |
| | Voltage, potential difference | $U$ | volt $= {\rm W}\cdot{\rm A}^{-1} = {\rm kg}\cdot{\rm m}^2\cdot{\rm s}^{-3}\cdot{\rm A}^{-1}$ | ${\rm V}$ | | |
| | Current | $I$ | ampere (base unit) | ${\rm A}$ | | |
| | Resistance | $R$ | ohm $= {\rm V}\cdot{\rm A}^{-1} = {\rm kg}\cdot{\rm m}^2\cdot{\rm s}^{-3}\cdot{\rm A}^{-2}$ | $\Omega$ | | |
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| Conventional current direction: current flows from positive to negative. | {{tag>inductors air_core_coil magnetic_field hall_sensor transient_response current_density chapter1_1}}{{include_n>300}} |
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| ==== Direct resistance measurement ==== | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Hall-Sensor Calibration Coil: Short Air-Core Coil #@TaskText_HTML@# |
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| Determine the nominal value and the measured value of the resistance of $R_1$ (brown, green, orange), $R_2$ (yellow, violet, red), $R_3$ (red, violet, red) and the incandescent lamp $R_{\rm L}$. Also measure the approximate resistance $R_{\rm K}$ of your body from your right hand to your left hand. | A Hall-sensor calibration bench uses a short air-core coil to create a defined magnetic field. An air-core coil is chosen because it avoids hysteresis and remanence effects. The coil is wound as a short cylindrical coil. |
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| ^ ^ $R_1$ ^ $R_2$ ^ $R_3$ ^ $R_{\rm L}$ ^ $R_{\rm K}$ ^ | Data: \begin{align*} l &= 22~{\rm mm} \\ d &= 20~{\rm mm} \\ d_{\rm Cu} &= 0.8~{\rm mm} \\ N &= 25 \\ \rho_{\rm Cu,20^\circ C} &= 0.0178~{\rm \Omega\,mm^2/m} \end{align*} |
| | Nominal value | | | | | | | |
| | Measured value | | | | | | | |
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| How do you explain the deviation between $R_{\rm L,nom}$ and $R_{\rm L,meas}$? | A DC current of $1~\rm A$ shall flow through the coil. |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| What consequences can $R_{\rm K}$ have? | 1. Calculate the coil resistance $R$ at room temperature. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~301~@# <WRAP leftalign> The wire cross section is \begin{align*} A_{\rm Cu} &= \pi\left(\frac{d_{\rm Cu}}{2}\right)^2 = \pi(0.4~{\rm mm})^2 \\ &= 0.503~{\rm mm^2} \end{align*} The total wire length is approximated by the number of turns times the circumference: \begin{align*} l_{\rm Cu} &= N\pi d \\ &= 25\pi \cdot 20~{\rm mm} \\ &= 1570.8~{\rm mm} = 1.571~{\rm m} \end{align*} Thus, \begin{align*} R &= \rho_{\rm Cu}\frac{l_{\rm Cu}}{A_{\rm Cu}} \\ &= 0.0178~{\rm \Omega\,mm^2/m}\cdot\frac{1.571~{\rm m}}{0.503~{\rm mm^2}} \\ &\approx 0.0556~{\rm \Omega} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~301~@# \begin{align*} R = 55.6~{\rm m\Omega} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Now also determine the series and parallel combinations of resistors $R_1$, $R_2$ and $R_3$. State the formulae used: | 2. Calculate the coil inductance $L$. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~302~@# <WRAP leftalign> For this short air-core coil, use \begin{align*} L = N^2 \cdot \frac{\mu_0 A}{l}\cdot\frac{1}{1+\frac{d}{2l}} \end{align*} with \begin{align*} A &= \pi\left(\frac{d}{2}\right)^2 = \pi(10~{\rm mm})^2 = 314.16~{\rm mm^2} = 3.1416\cdot 10^{-4}~{\rm m^2} \\ \mu_0 &= 4\pi\cdot 10^{-7}~{\rm Vs/(Am)} \end{align*} Therefore, \begin{align*} L &= 25^2 \cdot \frac{4\pi\cdot 10^{-7}\,{\rm Vs/(Am)} \cdot 3.1416\cdot 10^{-4}~{\rm m^2}}{22\cdot 10^{-3}~{\rm m}} \cdot \frac{1}{1+\frac{20}{2\cdot 22}} \\ &\approx 7.71\cdot 10^{-6}~{\rm H} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~302~@# \begin{align*} L = 7.71~{\rm \mu H} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| $R_{\rm series} = R_{\rm a} + R_{\rm b}$ | 3. Which DC voltage must be applied so that the stationary current becomes $I=1~\rm A$? How large is the current density $j$ in the copper wire? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~303~@# <WRAP leftalign> In the stationary DC state, the coil behaves like its ohmic resistance: \begin{align*} U &= RI \\ &= 55.6~{\rm m\Omega}\cdot 1~{\rm A} \\ &= 55.6~{\rm mV} \end{align*} The current density is \begin{align*} j &= \frac{I}{A_{\rm Cu}} \\ &= \frac{1~{\rm A}}{0.503~{\rm mm^2}} \\ &\approx 1.99~{\rm A/mm^2} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~303~@# \begin{align*} U = 55.6~{\rm mV} \\ j = 1.99~{\rm A/mm^2} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| $R_{\rm parallel} = (R_{\rm a} \parallel R_{\rm b}) = \frac{R_{\rm a} \cdot R_{\rm b}}{R_{\rm a} + R_{\rm b}}$ | 4. How much magnetic energy is stored in the coil in the stationary state? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~304~@# <WRAP leftalign> \begin{align*} W_m &= \frac{1}{2}LI^2 \\ &= \frac{1}{2}\cdot 7.71\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\ &= 3.86\cdot 10^{-6}~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~304~@# \begin{align*} W_m = 3.86\cdot 10^{-6}~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| ^ ^ $R_1 + R_2$ ^ $R_1 + R_3$ ^ $R_2 + R_3$ ^ $R_1 \parallel R_2$ ^ $R_1 \parallel R_3$ ^ $R_2 \parallel R_3$ ^ | 5. Give the time-dependent coil current $i(t)$ when the coil is switched on. <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~305~@# <WRAP leftalign> A coil current cannot jump instantly. It starts at $0$ and approaches the final value $I=1~\rm A$ exponentially: \begin{align*} i(t) = I\left(1-e^{-t/T}\right) \end{align*} So the sketch starts at $0~\rm A$, rises quickly, and then slowly approaches $1~\rm A$. </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~305~@# \begin{align*} i(t) = 1\left(1-e^{-t/T}\right)~{\rm A} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
| | Calculated | | | | | | | | |
| | Measured | | | | | | | | |
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| ==== Indirect resistance measurement ==== | 6. How long does it take until the current has practically reached its stationary value? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~306~@# <WRAP leftalign> The time constant is \begin{align*} T &= \frac{L}{R} = \frac{7.71~{\rm \mu H}}{55.6~{\rm m\Omega}} \\ &\approx 138.9~{\rm \mu s} \end{align*} A practical final value is reached after about $5T$: \begin{align*} t \approx 5T = 5\cdot 138.9~{\rm \mu s} \approx 695~{\rm \mu s} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~306~@# \begin{align*} t \approx 695~{\rm \mu s} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| Resistance can also be determined by a current/voltage measurement. | 7. How much energy is dissipated as heat in the coil resistance during the current build-up? <WRAP group> <WRAP half column rightalign> #@PathBegin_HTML~307~@# <WRAP leftalign> Using the current from task 5, \begin{align*} i(t) = I\left(1-e^{-t/T}\right) \end{align*} the heat dissipated in the winding resistance up to the practical final time $5T$ is \begin{align*} W_R &= \int_0^{5T} R\,i^2(t)\,dt \\ &= R I^2 \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \end{align*} For this interval, the integral is approximately \begin{align*} \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \approx \frac{7}{2}T \end{align*} Thus, \begin{align*} W_R &\approx RI^2\cdot \frac{7}{2}T \\ &= 0.0556~{\rm \Omega}\cdot (1~{\rm A})^2 \cdot \frac{7}{2}\cdot 138.9~{\rm \mu s} \\ &\approx 27.05\cdot 10^{-6}~{\rm Ws} \end{align*} </WRAP> #@PathEnd_HTML@# </WRAP> <WRAP half column> #@ResultBegin_HTML~307~@# \begin{align*} W_R \approx 27.05\cdot 10^{-6}~{\rm Ws} \end{align*} #@ResultEnd_HTML@# </WRAP> </WRAP> |
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| **Ohm's law:** In a circuit, the current increases with increasing voltage and decreases with increasing resistance. | #@TaskEnd_HTML@# |
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| $ I = \frac{U}{R} $ | |
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| Build the measurement circuit shown in Figure 2 for each of the three resistors and set the voltage on the bench power supply to $12 ~{\rm V}$. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_2_indirect_resistance_measurement.svg}} | |
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| Measure $U_n$ and $I_n$. From these values calculate $R_n$ in each case. | |
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| ^ $I_1 / {\rm mA}$ ^ $U_1 / {\rm V}$ ^ $R_1 / {\rm k}\Omega$ ^ $I_2 / {\rm mA}$ ^ $U_2 / {\rm V}$ ^ $R_2 / {\rm k}\Omega$ ^ $I_3 / {\rm mA}$ ^ $U_3 / {\rm V}$ ^ $R_3 / {\rm k}\Omega$ ^ | |
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| ==== Kirchhoff's voltage law (loop law) ==== | |
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| In every closed circuit and in every supply loop, the sum of all voltages is zero. | |
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| Set the voltage on the bench power supply to $12 ~{\rm V}$ and measure this voltage accurately with a multimeter. Build the measurement circuit shown in Figure 3. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_3_loop_law.svg}} | |
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| Complete the voltage arrows and measure $U$, $U_1$ and $U_2$. | |
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| ^ $U$ ^ $U_1$ ^ $U_2$ ^ | |
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| What is the loop equation here? | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Verify the formula using the measured values: | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| The resistors $R_1$ and $R_2$ connected in series form a voltage divider. In what ratio are the voltages $U_1$ and $U_2$? | |
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| $U_1 / U_2 =$ <wrap onlyprint> \\ \\ </wrap> $=$ <wrap onlyprint> \\ \\ </wrap> | |
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| ==== Kirchhoff's current law (node law) ==== | |
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| At every branch point, the sum of all currents flowing into and out of the node is zero. | |
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| Set the voltage on the bench power supply to $12 ~{\rm V}$ and measure the voltage accurately with a multimeter. As a first step, build the measurement circuit shown in Figure 4. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_4_branch_currents.svg}} | |
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| Draw the arrows for the directions of currents $I_1$ and $I_2$ in Figure 4. On both multimeters the DC current range and the polarity must be set before switching on. Then measure currents $I_1$ and $I_2$ and enter the measured values in Table 5. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_4_total_current_and_node_K.svg}} | |
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| In what ratio are currents $I_1$ and $I_2$? | |
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| $I_1 / I_2 =$ <wrap onlyprint> \\ \\ </wrap> $=$ <wrap onlyprint> \\ \\ </wrap> | |
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| Switch the bench power supply on again and measure the current $I$. Enter its value in Table 5. | |
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| ^ $I$ ^ $I_1$ ^ $I_2$ ^ | |
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| Determine the node equation for node $K$ and verify its validity. | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Using the measured values of resistors $R_1$, $R_2$ and $R_3$, calculate the total resistance $R_{\rm KP}$. | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Using the calculated value of $R_{\rm KP}$, verify the measured value of the total current: | |
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| $I = \frac{U}{R_{\rm KP}} =$ <wrap onlyprint> \\ \\ </wrap> $=$ <wrap onlyprint> \\ \\ </wrap> | |
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| ==== Voltage divider as a voltage source (a) ==== | |
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| The voltage divider shown in Figure 6 is initially in the unloaded condition, because the entire current supplied by the bench power supply flows through the series-connected resistors $R_1$ and $R_2$. A resistor connected in parallel with $R_2$ loads the voltage divider. | |
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| Set the voltage on the bench power supply to $12 ~{\rm V}$ and measure the exact voltage with a multimeter. Build the measurement circuit shown in Figure 6. | |
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| For the connected load $R_{\rm L} = 10 ~{\rm k}\Omega$, the voltage divider represents a voltage source. Like any voltage source, it has a source voltage (open-circuit voltage) $U_0$ and an internal resistance $R_{\rm i}$. The internal resistance of the voltage divider, regarded as a voltage source, results from the parallel connection of divider resistors $R_1$ and $R_2$: | |
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| $R_{\rm i} = R_1 \parallel R_2 = \frac{R_1 \cdot R_2}{R_1 + R_2}$ | |
| \\ | |
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| Using the measured values of resistors $R_1$ and $R_2$, calculate the internal resistance of the voltage source and determine the source voltage: | |
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| $R_{\rm i} =$ <wrap onlyprint> \\ \\ </wrap> \\ | |
| $U_0 =$ <wrap onlyprint> \\ \\ </wrap> | |
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| The power supplied by the bench power supply $P_0$ can be calculated using the following equation: | |
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| $P_0 = U \cdot I_1$ | |
| \\ | |
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| The power consumed by the load resistor can be determined using the following equation: | |
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| \\ | |
| $P_{\rm L} = R_{\rm L} \cdot I_2^2$ | |
| \\ | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_6_loaded_voltage_divider.svg}} | |
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| ==== Voltage divider as a voltage source (b) ==== | |
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| Draw the equivalent voltage source of the voltage divider: | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_6b_equivalent_voltage_source.svg}} | |
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| What value would $U_2$ have without $R_{\rm L}$? | |
| $U_{2,0} =$ <wrap onlyprint> \\ \\ </wrap> | |
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| Calculate $U_{2{\rm L}}$ and $I_2$ for $R_{\rm L} = 10 ~{\rm k}\Omega$ using the values of the equivalent voltage source. State the formulae used. | |
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| $U_{2{\rm L}}:$ <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| $I_2:$ <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Verify the values by measurement: | |
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| $U_{2{\rm L},meas}:$ <wrap onlyprint> \\ \\ </wrap> | |
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| $I_{2,{\rm meas}}:$ <wrap onlyprint> \\ \\ </wrap> | |
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| Verify the values using Kirchhoff's laws. State the formulae used. | |
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| $U_{2{\rm L}}:$ <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| $I_2:$ <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| ==== Non-linear resistors ==== | |
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| All resistors investigated so far are linear resistors, for which the characteristic $I = f(U)$ is a straight line. See Figure 7. The resistance value of a linear resistor is independent of the current $I$ flowing through it or of the applied voltage $U$. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_7_linear_characteristic.svg}} | |
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| For non-linear resistors there is no proportionality between current and voltage. The characteristic of such a resistor is shown in Figure 8. For these resistors one speaks of the static resistance $R$ and the dynamic (or differential) resistance $r$. | |
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| The static resistance is determined for a particular operating point: at a given voltage, the current is read from the resistance characteristic. The calculation is carried out according to Ohm's law: | |
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| $R = \frac{U}{I}$ | |
| \\ | |
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| The differential resistance around the operating point is calculated from the current difference caused by a change in the applied voltage: | |
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| $r = \frac{\Delta U}{\Delta I}$ | |
| \\ | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_8_non_linear_characteristic.svg}} | |
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| As an example of a non-linear resistor, an incandescent lamp is investigated. Build the measurement circuit shown in Figure 9. | |
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| {{drawio>lab_electrical_engineering:1_dc_circuit_theory:figure_9_incandescent_lamp_measurement_circuit.svg}} | |
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| Set the bench power supply to the voltage values from Table 7. Measure the corresponding current values and enter them in Table 7. | |
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| ^ $U / {\rm V}$ ^ 0.5 ^ 1.0 ^ 2.0 ^ 3.0 ^ 4.0 ^ 5.0 ^ 6.0 ^ 7.0 ^ 8.0 ^ | |
| | $I / {\rm mA}$ | | | | | | | | | | | |
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| Plot the characteristic $I = f(U)$. | |
| <wrap onlyprint> \\ \\ \\ \\ \\ \\ </wrap> | |
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| Calculate the static resistance $R$ at the operating point $U = 7.0 ~{\rm V}$. | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Calculate the dynamic resistance $r$ at the operating point $U = 7.0 ~{\rm V}$. | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |
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| Compare the values with those from Section 1.2 (direct resistance measurement). | |
| <wrap onlyprint> \\ \\ \\ \\ </wrap> | |