GOAL Calculate elementary properties cfa simple E circuit. PROBLE An uncharged c
ID: 1784322 • Letter: G
Question
GOAL Calculate elementary properties cfa simple E circuit. PROBLE An uncharged capator and a resistor are connected in series to a battery. If E 12.0 = 5.00 F, and R-8.OOx 105 12, find (a) the time constant of the circuit, (b) the maximum chargc on the capocitor, (c) the charge on the capocitraftcro.00 s. (d) thepotcntial dfferencc cro the resistor ater 6.00 s, and (e) therent in the resistort thst trne - STRATEGY Finding the time constant in part (a requires substitution into the timcor stant of an RC circuit. Far nat (h), tr maximum chara ocou attar long time, when tha current has iropped to 7-o. By orm's lau·dW-R, tha potential irference acrnes the-esistur is also cero dt that ti" ", dnd Kirchhoff's loop rule then givthee "igsrnur" dw inding the charge at soma particular time, as In part ic), is matter cf substituting into tne propar cquation. Kirchhoft's loop nle and tha canadtarca equation can b used to Indlrectty find tha potential drop across the resister in part id), and then dnm's law yiald, tha currant SOLUTTON (A) Find the time constant of the dircuit. use the dafinition of the time castnt.ERC103 0.0010F. (B) Calculate the maximum charg on the capacitor Apply Kinhhofrs loop n.le to the AC icuil, ing cduckwise, which mea battary is postive and tha ditrarancaa arrass th capacitar and rAGistor are nagotive Fromthe drinitaw or capactance and (2) -tra- Ohm's law, wa hava aveq and d-TR. These are votage dreps, so they're negative. Also, .bat = When the maximum chargs q- is rcached, 1-0. Sclvc Equation(2) for tha maximum charga Substitute to find the maximum charga -o Q-(5.00x 10 (12.0 V)-60.0 pc (C) Find the chargc on the capocitor after 6.00s Gubstitute into the time chere (U) Uompute thepotential diTerence scrces therewstor dfter .00 s. Compute the voltage drop Vcrcs the capacitor at that time Solve Equation 1) for avR dnd suhstitute. Fnd the currant in the realstor (E) pply Ou's lew, using the results o part (d) (remember that avR =-IR after G.00. .00 x 1U 12) KLMARKS Ln solvinu tos problem, vve peid scrupulous duertior, to signs. hest signs trust dlwdys b chosen when applying Kircthoffs lcop rule and must remain consistent throughout the problem. Alternately, magnitudes can te ued and the sigrs chosen by physical intuition. For example, the magnitude of the potential cifference across tha resistor mt eual the magniude of the patential difference across the battery minus the magritude of the potential difference across the capacitor QUESTION In an RC circuit aa dapicted in the tigure above, what happans to the tima requirad tor tha capacitor to h Charged to har its maeimum valueit ither th rasistance ar capatanc isincr ated with the sama appied vnltage? (select all thar apply Increasing the restance increases thetime. tncreasin th resistance ecraases the tima. Lntreesin tve capdldnce increases the time. Increasing tne capacitance decreases thc time. connected in series tu d battery,shown in the "uure dbove. R-7.10 x . find thc following. 14.0= 5.20, +, snd (a) the time constant t the cirat b) tha maximum changa on tha capacror (c) the charge on thc capadtor ortar 5.70 (d) the potential difference across the resistor after 5.7O 5 (e) the ourrent in the resistor at that time HINTS: GLTIING SIARILD ITMSICK Lisa the values tram PRACTICF IT ta heip yau work this ac (a) Hind th uldre on the capacitor dfter 2.10 s heve elapsed. XXertial difference across th-on-ato' after 2.10 s. (b)Tind the 'nagnitud-of th VE- c) Find the martude of the potential diterence acoss the resistor at that same time. Need Help? AuExplanation / Answer
Practice it:
E = 14 V
C = 5.2 uF
R = 7.4*10^5 ohm
a) Time constant = RC = 3.85 s
b) Maximum charge on the capacitor = Q = CV = 5.2 uF*14 = 72.8 uC
c) q = Q(1 – e^-t/RC) = 72.8*(1 – e^(-5.7/3.85)) = 56.2 uC
d) Voltage across the capacitor = -q/C = -56.2/5.2 = -10.8 V
Then voltage across resistor = -14 – (-10.8) = -3.2 V
e) Current in the resistor = I = 3.2/(7.4*10^5) = 4.32 uA
Exercise:
(a) q = Q(1 – e^-t/RC) = 72.8*(1 – e^(-2.1/3.85)) = 30.6 uC
(b) Voltage across the capacitor = -q/C = -30.6/5.2 = -5.88 V
(c) Voltage across resistor = -14 – (-5.88) = -8.12 V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.