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Unformatted text preview: Lab 6 RC Time Constant
'W Lab 6 Report 100 oints) Name: PHYS 1214 2114 Section: 3 Station Checkin: 3 z 1. Print a copy of your voltage vs. time grap
the corresponding circuit element as in F i PtS) hs. Label each trace (voltage vs. time graph) with
gure 6 and attach the copy to your lab report. (10 2. Use the smart tool and zoom options for the computer simulation to complete Table 1. Refer
to Figure 6 for details. Note that in Table l, as in Figure 6, Vs denotes the instantaneous voltage across the function generator, VR denotes the instantaneous voltage across the 10052 resistor, Vc denotes the instantaneous voltage across the lOOuF capacitor, x t1 is sometime shortly after the voltage is applied (i.e., the switch is closed), t; is the instant of time at which the voltage drop across the 1009 resistor is equal to the
potential difference across the lOOuF capacitor, ' t3 is the instant of time at which the voltag e across the IOOuF capacitor has reached 63%
of its maximum value, and t4 is sometime shortly before the applied voltage is turned off and the lOOuF capacitor
begins to discharge. Table 1. RC Series Circuit Instantaneous Volta  e Data 40 ts) VR . . Lab 6 RC Time Constant How is the direction of the current ﬂow as the capacitor discharges related to the direction of
the current ﬂow as the capacitor charges? Explain your reasoning. (Hint: How is the
direction of current ﬂow through a resistor related to the sign of the potential difference
across the resistor?) (8 pts) As 71% 5,35,41,92 3m” is 94,39??? Ma cut/M 7g ﬂatly 742in #1
cafuilow (fofrkvgjajdol when H's“ d;<;:::% ﬁfth; r+ TS 9057 “W67 Vctrow ﬁt
Ca “Flay (whim) , 71¢ 9‘7,“ 0“; Cuwemt— How W [a Maﬁa] 0/;Mewemc e
z” m M 9m 6. Show that the quantity RC has units of time, that is, the product of ohms and farads reduces to seconds. (Recall that l ampere of current is equal to 1 coulomb of charge per second.) (8 13125) fit—.20" (Cempz—fpaéf,§:$‘
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ﬁCa5 7. Calculate the sum of the voltage drop VR across the 1009 resistor and the potential
difference Vc across the lOOuF capacitor for each of the times listed in Table 1. Record the
calculation as VR + Vc in Table 1. According to Kirchhoff’ 8 Loop Rule, at each instant of x» time, the sum of voltage drop across the resistor and the potential difference across the
capacitor is equal to the voltage Vs applied by the function generator. How well does
Kirchhoff’ s Loop Rule hold for the RC series circuit studied in lab? (4 pts) ’YQ/g) WV” QKF¢v2W vast/{W K7,.» ch!) 7c??? LO??? ”Wife W47 Wel/ (W +Vc=V‘7) On the graph below, accurately sketch the voltage across the resistor in an RC circuit for the
square wave output of a function generator shown below. The output of the function
generator is indicated by a solid line in the graph and the potential difference across the
capacitor is indicated by a dashed curve. (Hint: Kirchhoff’ s Loop Rule.) (6 pts) 10/11/2011 07:01 AM Pasco Lab 6.ds Scope 1 (0.3921. 3.973) §
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w Lab 6 RC Time Constant C‘ 3. The experimental time constant, rexp, for the RC circuit i ﬁglﬁfﬂgﬁ _ een the time t1
at which the voltage was applied and the time t3 whélg; vgtage g f, ' _ the capacitor reached 63% of its maximum value. (12 pts) rexp=t3—t1= omﬁw ~ 0.33m. '3 20'5”? 3; How does the experimental time constant compare to the theoretical time constant I = RC?
Show all work, including correct use of units. 1=RC= QooJIIXmemwﬁ); 5.0;??? / 0‘03 ’ 9,0415]
%error= Vx/oey, ”:1 1’3}: 4. Compare the time dependent behavior of the voltage across the IOOuF capacitor and the
voltage across the 1009 resistor observed in lab (that is, represented by the voltage vs. time
graphs generated in lab) to the theoretical time dependent behavior expressed in Equations 6
and 7 for charging a capacitor. (8 pts) Capacitor 723., zymfh we; gmmw aggggw Vi. r... £(/__ EL“ $31.2“) M \p
1/0“ W Jo,“ W H {3 alarm}. Damage, way: ,‘3
at. «gym ewifﬁ i galvaﬁi’t Resistor ,‘f
[4‘2 ﬂaw/SF‘i‘iﬁ’V aﬂD/l/Wf ﬂab Wrﬁ ~£1Cﬂi éwmlq, 17:96 4:: , a/ J W14 W Mg; Magi it? vlma éﬁnmiig me “a exam x” r gmjii’é" if! 004:5an M 5. The time dependence of the voltage across the resistor as the capacitor discharges (cf. Figure
6) can be explained using the deﬁnition of capacitance and Kirchhoff’s Loop Rule: hag. saiawaa mm? As the capacitor discharges, the instantaneous voltage across the capacitor decreases
(exponentially) to 0 volts (C V = Q, Qi => Vcl) In accordance with Kirchhost
Loop Rule, the instantaneous voltage across the resistor is negative and increases
(asymptotically) to 0 volts (V s = 0 = VR + Vc, VCiO => VR < 0 and VRTO). In
accordance with Ohm’s Law, the magnitude of the current ﬂowing through the circuit
as the capacitor discharges decreases to 0 amps. Comment as to whether this is consistent with the time dependent behavior of the voltage
across the 1009 resistor that you observed in lab (that is, represented by the voltage vs. time graphs generated in lab). (4 pts) ﬂ ...
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 Spring '08
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