## Transcribed Text

8.1
Charging a capacitor
Figure 1: The positive and negative terminals of a capacitor.
the Ir: lectures, we found that the voltage across a capacitor is given by Vacross cap = 8 where
in charge in coulombs on one plate of the capacitor and C is the capacitance of the capacitor is
farads. From this relationship, we cau see that as we put more charge onto the plates of
capacitor, the voltage across the capacitor will increase (and vice versa if we remove charge from
a
the plates of a capacitor).
Consider the following circuit:
-q
/
Switch
Vo
les
Figure 2: Charging a capacitor. Catneli, J. D. and K. W Johnson (2012). Physics. Hobaken,
NJ, Wiley.
capacitor has no charge on it initial'y Because thers is no excess charge will on flow the
Say that the the voltage V across the capacitor is vero (*). When the switch is closed charge
plates,
271
and
Increase with on time(* the capacitor Because plates. This will mean that the voltage across the capacitor will
collect
in the citeuit is limited this of the resistor in the circuit, the amount of charge that can flow
it repels capacitor increases slowly(*). As builds up on the plates,
the voltage across the means that the charge on the capacitor builds up slowly and heuce
across be deposited incoming capacitor on the charge capacitor will with plates bigger and time bigger force charge This means that less and capacitor less charge
will
the as goes by. Therefore, the rate of increase of voltage
also why our capacitor computers that explains the slow propagation of nerve signals in the human body and the
plates of a decrease as time goes by. It is this slow increase of voltage across
over
this in lecture are limited in how fast they can compute if we have time we will go
It
charged can be the shown voltage (see across your text the capacitor for the derivation) varies as: that when an initially uncharged capacitor
is
(1)
obms) the resistor, C is the capacitance (in farads) of the capacitor, and t is the time the
whete V of is the voltage across the capacitor, Vo is the battery voltage, R is the resistance (in
v.S. capacitor has been charged for in the circuit. If we plot a graph of voltage across the capacitor
time the capacitor has been charged for, we get the plot in Figure 3.
Time (t)
Figure 3: Charging a capacitor.
Exercise 2 Suppose Vo = 9 volts, R = 270 ks. C = 820 uF in Figure 2, calculate the voltage
accoss the capacitor, 37 seconds after the switch is closed
Notice that if we wait a time t = RC, the voltage across the capacitor is:
V = Vo(1 RE) Vo(1 = Vo(1 0.368) - 0.63V6
i.e. the voltage across the capacitor has increased to 63% of Vo after t RC seconds This time
is called the time constant of the circuit The time constant of a circuit is useful as it gives you
a quick idea of how fast voltages can change in a circuit.
Exercise 3: Suppose Vo = 9 volts, R = C = 820 F in Figure 2, calculate the time
constant of the circuit.
Exercise 5: Suppose R = 270k9, C = 820 uF in Figure 4. calculate the voltage across the
capacitor, 37 seconds after the resistor is connected across the charged capacitor. The capacitor
is initially charged to 9 volts.
Notice that if we wait a time = RC after the resistor has been placed across the capacitor, then
the voltage across the capacitor is:
V6(0.368) R
i.e. the voltage across the capacitor has decreased by 63% of Vo after t = RC seconds. This time
is called the time constant T of the circuit. The time constant of a circuit is useful as it gives you
a quick idea of how fast voltages can change in a circuit.
Exercise 6: Suppose R = 270 kn C = 820 uF in Figure 4, calculate the time constant of the
circuit. The capacitor is initially charged to 9 volts.

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