(a) A capacitor in series with a resistor, switch, and battery. (b) When the swi
ID: 1419170 • Letter: #
Question
(a) A capacitor in series with a resistor, switch, and battery. (b) When the switch is thrown to position a, the capacitor begins to charge up. (c) When the switch is thrown to position b, the capacitor discharges. An uncharged capacitor and a resistor are connected in series to a battery as shown in the figure, where e m f = 12.0 V, C = 4.80 µF, and R = 7.1 X10^5 . The switch is thrown to position a. Find the charge and current as functions of time. Conceptualize: Study the figure and imagine throwing the switch to position a as shown in Figure (b). Upon doing so, the capacitor begins to charge. Categorize: We evaluate our results using equations developed in this section, so we categorize this example as a substitution problem. The time constant of the circuit is 3.408s, the maximum charge on the capacitor 57.6microC, and the maximum current in the circuit is 16.9microA. Find the charge and current as functions of time (in seconds): Use "t" as necessary.
Explanation / Answer
consider a circuit as shown in diagram when the key is connected to point a the charging of capacitor takes place
until the potential difference across the plates of the condenser becomes E
but charge attained already on the plates opposes further introduction of charge
E-q/c=Ri
[Ec-q]/c=Rdq/dt
dq/[Ec-q]=dt/cR
we applying intergration on both side we get
log[Ec-q]-log[Ec]=t/cR
log{[Ec-q/Ec]=t/cR
Ec-q/Ec=e^-t/cR
1-q/Ec=e^-t/cR
q/Ec=1-e^-t/cR
q=Ec{1-e^-t/cR}
q=qo{1-e^-t/cR}
this equation for charge as function of time
the current i=dq/dt=d/dt{qo[1-e^-t/cR]}=ioe^-t/cR
(b) ans
time constant T=cR=3.408 s
maximum charge qo=57.6*10^-6 C
maximum current io=16.9*10^-6 A
now we find the charge
charge q=57.6*10^-6{1-e^-1/3.408}=14.93*10^-6 C
the current I=16.9*10^-6*e^-1/3.408=12.52*10^-6 A
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