Soal#1
A 4.6 µF capacitor that is initially uncharged is connected in series with a 7.50 kΩ resistor and an emf source with ε = 245 V and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts (a)–(d)?Answer:
Known:
Capacitance, C = 4.6 µC
Resistor, R = 7.50 kΩ = 7500 Ω
emf source with ε = 245 V
The voltage across a capacitor is VC = q/C
(a) At the instant the circuit is completed, there is no voltage across the capacitor, since it has no charge stored.
(b) Since VC = 0. The full battery voltage appears across the resistor VR = ε = 245 V
(c) There is no charge on the capacitor.
(d) the current through the resistor is i = ε/Rtotal = 245 V/7500 Ω = 0.0327 A = 32,7 mA
(e) After a long time has passed the full battery voltage is across the capacitor and i = 0. The voltage
across the capacitor balances the emf:
VC = 245 V
The voltage across the resistor is zero. The capacitor’s charge is
q = CVC = (4.60 x 10−6 F)(245 V) =1.13 x 10−3 C. The current in the circuit is zero.
Problem#2
A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit?
Answer:
The capacitor discharges exponentially through the voltmeter. Since the potential difference across the capacitor is directly proportional to the charge on the plates, the voltage across the plates decreases exponentially with the same time constant as the charge.
The reading of the voltmeter obeys the equation V = V0e─t/RC
where RC is the time constant.
(a) Solving for C and evaluating the result when t = 4.00 s gives
C = t/[R ln (V/V0)]
C = 4.00 s/[3.40 x 10─6 Ω ln (12.0 V/3.0 V)]
C = 8.49 x 10−7 F = 0.849 μF
(b) the time constant of the circuit is
tc = RC = (3.40 x 106 Ω)(8.49 x 10−7 F) = 2.89 s
Problem#3
A capacitor is connected through a 0.895 MΩ resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants, and (c) Graph the results of parts (a) and (b) for t between 0 and 20 s.
Answer:
Known:
Resistor, R = 0.895 MΩ = 8.95 x 105 Ω
potential difference, V = 60.0 V
The time constant is RC = (8.95 x 105 Ω)(12.4 x 10−6 F) = 11.1 s
(a) the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s is
At t = 0 s: q = Cε (1− e−t/RC) = 0
At t = 5 s: q = Cε (1− e−t/RC) = (12.4.10−6 F)(60.0 V)(1− e−(5.0 s)/(11.1 s)) = 2.70.10−4 C.
At t =10 s: q = Cε (1− e−t/RC) = (12.4.10−6 F)(60.0 V)(1− e−(10.0 s)/(11.1 s)) = 4.42.10−4 C.
At t = 20 s : q = Cε (1− e−t/RC) = (12.4.10−6 F)(60.0 V)(1− e−(20.0 s)/(11.1 s)) = 6.21.10−4 C.
At t =100 s : q = Cε (1− e−t/RC) = (12.4.10−6 F)(60.0 V)(1− e−(100 s)/(11.1 s)) = 7.44.10−4 C.
(b) the charging currents at the same instants is
The current at time t is given by:
i = (ε/R)e−t/RC
so on,
At t = 0 s: i = (ε/R)e−t/RC = (60.0 V/8.95 x 105 Ω)e−(0/11.1 S) = 6.70 x 10─5 A
At t = 5 s: i = (ε/R)e−t/RC = (60.0 V/8.95 x 105 Ω)e−(5 s/11.1 s) = 4.27 x 10─5 A
At t =10 s: i = (ε/R)e−t/RC = (60.0 V/8.95 x 105 Ω)e−(10 s/11.1 s) = 2.72 x 10─5 A
At t = 20 s : i = (ε/R)e−t/RC = (60.0 V/8.95 x 105 Ω)e−(20 s/11.1 s) = 1.11 x 10─5 A
At t =100 s : i = (ε/R)e−t/RC = (60.0 V/8.95 x 105 Ω)e−(100 s/11.1 s) = 8.20 x 10─5 A
(c) The graphs of q(t) and i(t) are given in Figure 26.44a and b.
 |
| Fig.1a |
 |
| Fig.2 |
In the circuit shown in Fig. 2 both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V, and (b) what will be the current at that time?
 |
| Fig.2 |
Known:
both capacitors are initially charged, V0 = 45.0 Vpotential across each capacitor is V = 10.0 V
The capacitors, which are in parallel, will discharge exponentially through the resistors. Since V is proportional to Q, V must obey the same exponential equation as Q,
V = V0 e−t/RC.
The current is
i = (V0/R)e−t/RC
(a) Solve for time when the potential across each capacitor is 10.0 V:
t = −RC ln(V/V0)
= –(80.0 Ω)(35.0 μF) ln(10 V/45 V)
t = 4210 μs = 4.21 ms
(b) Using the above values, with V0 = 45.0 V, gives
i = (45.0 V/80.0 Ω)e–4210 μs/(80.0 Ω)(35.0 μF)
i = 0.125 A.
Post a Comment for "R-C Circuits Problems and Solution"