Transients in RC and RL Circuits 0. Introduction The objective of this experiment is to study the DC transient behaviors of RC and RL circuits. This experiment has divided into 6 parts: 1. Charging curve from measured data ( R = 10M ? and C = 4 mF ) 2. Draw the charging curve by the graphical method 3. Discharging curve from measured data ( R = 5M ? and C = 4 mF ) 4. Draw the discharging curve by the graphical method 5. Display of the charging and discharging curve of capacitor 6. Display of the charging and discharging curve of inductor 1.
Theories (a) Capacitor
Capacitor is an electrical passive device for storing charge in the form of electric field. In its simplest from, It consists basically consists of two conductors which are separated by a dielectric medium (non-conductor) such as air, waxed paper, plastics, etc. The capacitance of capacitor is directly proportional to the surface areas and the inverse of the separation of the two conductors. The dielectric constant of the non-conductor is also affecting the capacitance. FIGURE 1 Capacitor symbol For an ideal capacitor, the capacitor current iC is proportional to the time rate of change of the voltage across the capacitor:
Where C is the proportionality constant and is known as capacitance.
(b) Inductor Inductor is an electrical passive device for storing energy in the form of magnetic field. In its simplest from, It consists basically consists of a wire loop or coil. The inductance is directly proportional to the number of turns in the coil. Inductance also depends on the radius of the coil and on the type of material around which the coil is wound. FIGURE 2 Inductor symbol For an ideal inductor, the inductor voltage VL is proportional to the time rate of change of the current through the inductor:
Where L is the proportionality constant and is known as inductance. (c) RC circuit RC circuit is consists of resistor and capacitor. The simplest form is shown in below. FIGURE 3 Simplest RC circuit For discharging case, when t<0 , then VS=0 By Kirchhoff’s Voltage Law, at the steady state, 0=VC-VR 0=IR-QC Assume the resistor and the capacitor are ideal (i. e. R and C are constant). Then we have 0=-RdQdt-QC (I=-dQdt) dQdt=-QRC QoQ1QdQ=0t-1RCdt lnQQo=-tRC, Q=Qoe-tRC By VC=QC , VC=QoCe-tRC VC=Voe-t? Where time constant, ? =RC For half life,
Vo2=Voe-t12? t12=ln2?? =0. 693? When t=? , VC=Voe-1 VC=0. 368Vo From the above, the half-life of capacitor voltage was related to the time constant ? , when the time equal to t = 0. 693? , then the voltage remains half. And if the time equal to time constant, the capacitor voltage would decrease to 0. 368Vo For charging case, when t<0 , then VS=VO By Kirchhoff’s Voltage Law, at the steady state, 0=VC-VR 0=IR-QC Assume the resistor and the capacitor are ideal (i. e. R and C are constant). Then we have 0=-RdQdt-QC (I=-dQdt) dQdt=-QRC QoQ-Qo1QdQ=0t-1RCdt lnQ-QoQo=-tRC Q=Qo(1-e-tRC) By VC=QC ,
VC=QoC(1-e-tRC) VC=Vo(1-e-t? ) Where time constant, ? =RC For half life, Vo2=Vo(1-e-t12? ) t12=ln2?? =0. 693? When t=? , VC=Vo(1-e-1) VC=0. 632Vo From the above, the half-life of capacitor voltage was related to the time constant ? , when the time equal to t = 0. 693? , then the voltage remains half. And if the time equal to time constant, the capacitor voltage would decrease to 0. 632Vo (d) RL circuit RL circuit is consists of resistor and inductor. The simplest form is shown in below. FIGURE 4 Simplest RL circuit For current-out case, when t<0 , then VS=0
By Kirchhoff’s Voltage Law, at the steady state, 0=VL-VR 0=-LdIdt-IR Assume the resistor and the inductor are ideal (i. e. L and C are constant). Then we have IoI1IdI=0t-RLdt lnIIo=-tLR, I=Ioe-tLR I=Ioe-t? Where time constant, ? =LR For half life, Io2=Ioe-t12? t12=ln2?? =0. 693? When t=? , I=Ioe-1 I=0. 368Io From the above, the half-life of inductor current was related to the time constant ? , when the time equal to t = 0. 693? , then the current remains half. And if the time equal to time constant, the inductor current would decrease to 0. 68Io For current-out case, when t<0 , then VS=0 By Kirchhoff’s Voltage Law, at the steady state, 0=VL-VR 0=-LdIdt-IR Assume the resistor and the inductor are ideal (i. e. L and C are constant). Then we have IoI-Io1IdI=0t-RLdt lnI-IoIo=-tLR, I=Io(1-e-tLR) I=Io(1-e-t? ) Where time constant, ? =LR For half life, Io2=Io(1-e-t12? ) t12=ln2?? =0. 693? When t=? , I=Io(1-e-1) I=0. 632Io From the above, the half-life of inductor current was related to the time constant ? , when the time equal to t = 0. 693? then the current remains half. And if the time equal to time constant, the inductor current would decrease to 0. 632Io (e) Graphical method of drawing a charging curve. To draw a charging curve, two parameters are needed. 1. The final voltage Vm, which is the final value after the capacitor being fully charged. 2. The time constant ? = RC. Step 1. Draw an initial point A at the origin (t = 0 , v = 0) Step 2. Locate a point A’ on the line v = Vm such that A’ lags behind A by a time given by?. Step 3. Draw the line AA’. FIGURE 5 Graphical method Step 3 Step 4. Select a point B on line AA’.
If B were chosen at a point close to A, the resultant curve will be a better approximation to the actual charging curve (in the expense of more iterative steps). Step 5. Draw a line BB’ in the same fashion of line AA’. FIGURE 6 Graphical method Step 5 Step 6. Repeat the steps above for C, D, E, F, and so on. The line ABCDEF…. is the approximation to the actual charging curve. 2. Apparatus 1. Experimental board 2. Square wave generator 3. Oscilloscope 4. Digital voltmeter 5. DC power supply 6. Stop watch 3. Procedure 1. Charging curve from measured data (a) Digital voltmeter V was used to measure the supply voltage VS.
The supply voltage was VS recorded. (b) The digital voltmeter V in (a) was used and the circuit was connected as shown in FIGURE 9, with Vs=5V, R = 10 M? , C = 4 ? F, and the Switch SW to position FIGURE 8 RC circuit with R = 10 M? , C = 4 ? F (c) With the stop watch at “ready”, SW was switched from position 2 to position 1 and at the same instant the stop watch was started. The capacitor voltage was recorded at suitable intervals (at least 8) until the capacitor was fully charged, as shown by a steady maximum voltage. (d) The charging curve (volts on the vertical axis) was plotted on a graph paper. 2.
The charging curve was drawn by the graphical method (no measurement was needed). The circuit in FIGURE 9 could be represented by its equivalent circuit shown in Figure 10, where, RC is the equivalent resistance of R in procedure 1 and the internal resistance, r, of the voltmeter. The internal resistance of the DVM = 10 M? , so RC=R //r=10M? //10M? =5M? The Equivalent charging voltage=V? rr+R =5? 10? 10610? 106+10? 106 =2. 5 V FIGURE 9 The equivalent circuit of FIGURE 9 Draw the curve on the same graph paper as in step 1 (d). 3. Charging curve from measured data (a) Connect the circuit as shown in FIGURE 9, with R = 5 M? by connecting two 10 M? resistors in parallel) and C = 4 ? F. Switch SW to position 1. Wait for three minutes until the capacitor is fully charged. (b) The equivalent charging voltage was calculated. (c) The switch was switched from position 1 to position 2 and at the same instant when the stop watch started. (d) The capacitor voltage was recorded at suitable intervals to plot the decay curve of the capacitor voltage. 4. Graphical construction of discharge curve On the same graph paper as in step 3(d), the discharge voltage curve was plotted using the graphical method. 5.
Display of the charging and discharging curve (a) A circuit as shown on FIGURE 11. was connected. (b) Use FIGURE 11, a 300-Hz square wave input (2 volt peak to peak) and the waveform observed on the oscilloscope was recorded. FIGURE 10 The connection of CRO and RC circuit 6. Display of the charging and discharging curve (a) Circuit diagram FIGURE 11 The connection of CRO and RL circuit (b) When a step voltage V was applied across an RL series circuit the inductor current may be obtained by the relation: (c) Use the circuit above and record the waveform of the inductor voltage, VL, as shown on the CRO. d) The curve of the resistor voltage (iLR) was obtained by subtracting the VL curve from the DC step value V, hence the current growth in the inductor due to the step voltage V was determined. 4. Measurements Procedure 1, (a) The voltage supply measured is 5. 05V, time constant is 39. 5s (c) The taken of time intervals are 5 second Time (s)| Voltage (v)| 0| 0. 00| 5| 0. 68| 10| 0. 972| 15| 1. 28| 20| 1. 55| 25| 1. 78| 30| 1. 93| 35| 2. 06| 40| 2. 15| 45| 2. 22| 50| 2. 27| 55| 2. 32| 60| 2. 37| 65| 2. 39| 70| 2. 42| 75| 2. 43| 80| 2. 45| 85| 2. 46| (d) FIGURE 12 Capacitor charging voltage against time (measured)
Procedure 2 FIGURE 13 Capacitor charging voltage against time (graphical) Procedure 3 (a) R=5M? , r=DVM=10M? RL=5M? //10M? =103M? (b) The equivalent charging voltage=V? (rr+R) =5. 05? 10? 10610? 106+5? 106 =3. 35V (c) The taken of time intervals are 5 second Time (s)| Voltage (v)| 0| 3. 35| 5| 2. 58| 10| 1. 78| 15| 1. 26| 20| 0. 80| 25| 0. 55| 30| 0. 41| 35| 0. 27| 40| 0. 20| 45| 0. 14| 50| 0. 10| 55| 0. 07| 60| 0. 05| (d) FIGURE 14 Capacitor discharging voltage against time (measured) Procedure 4 Time constant = 19715s FIGURE 15 Capacitor discharging voltage against time (graphical)
Procedure 5 The waveform picture of the Oscilloscope| | This is the waveform of capacitor voltage against the time. The input voltage is 300Hz square wave. | Procedure 6 | This is the waveform of inductor voltage against time. The waveform is generated by the inductor | Measured inductance of the inductor=0. 065mH Measured resistance of the resistor= 98. 97? The measured value of time constant =L/R=0. 065/98. 97=6. 57×10^-4s 5. Discussion 1. In FIGURE 9, why Rc = 5 M? and the charging voltage = 2. 5V? Use an appropriate theorem to explain this? FIGURE 16 Step of Thevenin resistance of FIGURE 10
FIGURE 17 Thevenin equivalence of FIGURE10 Since the resistance of the resistor is similar to the internal resistance of the digital voltmeter, the internal resistance could absorb half of the voltage. So that the voltage which supply by the voltage supplied to the capacitor is smaller, which means the final charged-up voltage of the capacitor is becoming smaller. FIGURE 16 shown the step of Thevenin resistance of the FIGURE 9. As considering the Thevenin resistance, the voltage source was treated as short-circuit, therefore the above figure was shown. RC=10. 025M? //10M? =5M? By Thevenin voltage law, Vab-510M+Vab10M=0 Vab=5 Vab=2. 5V Therefore the Thevenin equivalent circuit is consist with Vab=2. 5V and RC=5M? , which is shown in FIGURE 17. 2. Explain any differences between the experimental curves and the curves displayed on the CRO. For part 1, the differences of the experimental curves and the CRO curves is occurred in nearly time ? . The experimental curve is slightly higher than the CRO curves at time ? to time 3?. It may cause by: 1. The non-ideal of resistor and capacitor. As the actual case of resistor and capacitor, when the temperature increase, the reactance of the resistor and the capacitor must be higher.
This affect the time constant, as the time constant is proportional to the resistance, the time constant should be increased and the time required for complete charging is longer in general. Therefore, the charging rate in the actual case should be lower than the ideal case, so the experimental curve should slightly higher than the CRO curve if the other factors are fixed. 2. The human reaction time. As the time interval is taken as 5s, human reaction time is relatively large enough to affect the result. Since human reaction time is in the range of 0. 2 to 0. 4 second, the error could be as large as 0. 8%.
Because we are taken 18 sets of data, the maximum error could be ±14. 4%. For part 2, the differences of the experimental curves and the CRO curves is occurred when time is nearly equal to ? . The experimental curve is almost the same as the CRO curve. However, it is not perfectly match. The reason may be the same as that mentioned above. 3. Comment on the value of RC in (2) and the time taken (from the experiment curve) to reach 63. 2% of the total change. From Thevenin theorem, the value of thevenin resistance value RC is equal to R//r, that is 103M?. As the RC=403M? , therefore the time constant ? =RCC=103M? ? =403s. As the discharging voltage is: VC=Vo(1-e-t? ) If VC=0. 632Vo VC=Vo(1-e-1) ? t=? =RCC=403s The value of time taken to reach 63. 2% of the total change is about 15s, which is larger than the theoretical time constant. This may be due to the error in human reaction time during the recording of the experimental result. 4. Were the experimental results of (6) are expected. Explain. The result in (6) is expected. The ideal time constant=L/R=6. 57×10^-4s. From the experiment result, the time constant is only 1. 5? s. This is because the resistance of both the signal generator and the CRO is very high.
Therefore, the equivalence resistance of the circuit increased. Time constant=L/R, when the R increased, the time constant decreased. Therefore, as the resistance of the signal generator and CRO is very high, the time constant became very small. | 5. Comment on your charge and discharge curves. As the FIGURE 13 and FIGURE 15 are drawed by the best-fit line, the results are very near the theorem results. But it stills existing a little amount of error. Firstly, the experimental error had affected the resultant curve, for instance, the actual reactance of resistor, capacitor and inductor.
The humanical error is also being a huge problem for the result since it could cause a maximum of 14. 4% error. On the other hand, the graphical method requires as small as the time interval taken. As the time interval taken could not be infinitely small, the error due the graph always exists. Therefore the charging and discharging curves are not perfectly matching with the theorem curves as well as the graphical method curve. 6. Conclusion Since the experiment is success, a well study of the feature of capacitor, inductor, RC circuit and RL circuit was made.
The key concepts of this experiment are: 1. The time constant ? is RC in RC circuit and is R/L in RL circuit. 2. When the time equal to time constant, 63. 2% of voltage are charged or discharged. 3. The half life time of charging or discharging curves are equal to ln2??. 4. The resistance of the resistor could make a huge effect for the whole RC or RL circuit; therefore a calculation of equivalent circuit is advised. 5. The total voltage should equal to the superposition of resistor voltage and capacitor voltage. 6. The reaction time should count in the experimental result. 7.
As much as the experiment sets taken is advised, which could provide a much more accuracy result. 7. Reference  Capacitor http://en. wikipedia. org/wiki/Capacitor  Inductor http://en. wikipedia. org/wiki/inductor  P. Fung, P. Sun and K. Young. Further Physics, 3rd ed. Vol 2. Longman, 1996. The Physics Hypertextbook: Capacitance, 2009 The Physics Hypertextbook: Inductance, 2009. The Physics Hypertextbook: LC circuit, 2009 T. M. Mok, C. S. Wong & L. Y. Poh. New Way Phsics for Advance Level, Fields, Electricty and Electromagnetism. MANHATTAN PRESS (H. K. ) LTD, 2005.
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