Physics
Capacitors Solutions
+ Charging and Discharging a Capacitor in an R-C Circuit
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Learning Goal:
To understand the dynamics of a series R-C circuit.
Consider a series circuit containing a resistor of resistance R and a
capacitor of capacitance C connected to a source of EMF E with
negligible internal resistance. The wires are also assumed to have
zero resistance. Initially, the switch is open and the capacitor
discharged. (Figure 1)
Now that we have a feel for the state of the circuit in its steady state, let us obtain expressions for the charge of the capacitor and the current in the resistor as functions of
(
and It)
dg(t)
Using these equations, we obtain
da E
9
da(t)
de
dt
dt
R
and then,
RC
9(0)-CE
RC
time. We start with the loop rule: E – Vr - Vc = 0. Note that Vr(t) = f(t)r. Vo(t) = 940
Let us try to understand the processes that take place after the switch
is closed. The charge of the capacitor, the current in the circuit, and,
correspondingly, the voltages across the resistor and the capacitor, will
be changing. Note that at any moment in time during the life of our
circuit, Kirchhoffs loop rule holds and indeed, it is helpful:
E - VR - Vc = 0, where VR is the voltage across the resistor, and
Vc is the voltage across the capacitor.
y
Part 1
di
Integrate both sides of the equation
dg(0)
9C-CE
Rc to obtain an expression for q (t)
Express your answer in terms of any or all of E, R, t, and C. Enter exp(x) fore".
► View Available Hint(s)
VAZO
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9(t) =
Figure
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R
Part 1
fo=0
Now differentiate (t) to obtain an expression for the current I (t).
Express your answer in terms of any or all of E. R. t, and C. Enter exp(x) for e.
C=9=0
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