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Rc Circuit

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User Generated
Subject
Physics
School
University of California Irvine
Type
Homework
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The three capacitors are attached in parallel, so they may be treated as a single capacitor with an “equivalent capacitance” equal to the
sum of the three individual capacitances: C = 19 + 19 + 19 = 57 mF.
The resulting circuit just has three components: a battery, a resistor, and a capacitor. Kirchhoff’s Loop Rule then says:
V
battery
- IR - Q/C = 0 (since the voltage drop across the resistor is IR and across the capacitor is Q/C).
Since the current is just the rate of charge flow, we can write I = dQ/dt and solve the differential equation
V - R - Q/C = 0 by using the ansatz (guess) Q(t) = Q
0
e
λt
+ k. Plugging in this ansatz tells us that
V - RQ
0
λe
λt
- (Q
0
e
λt
+ k)/C = 0, and for this to hold true at all times, we must require that V - k/C = 0 (the constants cancel out) and
-RQ
0
λe
λt
- Q
0
e
λt
/C = 0 (the time-dependent parts cancel out). The solutions to these equations are k = CV and λ = -1/(RC). This gives us
Q(t) = Q
0
e
-t/(RC)
+ CV. To find Q
0
, we will require that Q(t=0) = 0, so Q
0
+ CV = 0 and hence Q
0
= -CV. Finally, we end up with
Q(t) = CV(1 - e
-t/(RC)
). Taking the derivative with respect to time will give us the current:
I(t) = CV· (1/(RC)· e
-t/(RC)
) = V/R· e
-t/(RC)
.
In order for the current to decrease to 10% of its initial value, we need to have e
-t/(RC)
= 1/10, which requires t = -RC· ln(1/10). Plugging
in the known values of R = 28Ω and C = 57×10
-3
F, we get t = 3.7 seconds.

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The three capacitors are attached in parallel, so they may be treated as a single capacitor with an “equivalent capacitance” equal to the sum of the three individual capacitances: C = 19 + 19 + 19 = 57 mF. The resulting circuit just has three components: a battery, a resistor, and a capacitor. K ...
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I was having a hard time with this subject, and this was a great help.

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