(Q5) Draw a circuit where a 6V DC voltage source is
connected in series with a switch, a 1kΩ resistor and a
30pF capacitor (initially uncharged). If we close the
switch at time t = 0, approximately how long will it take
for the voltage across the capacitor to reach 4V?
What is the voltage across the resistor at that time?
• This question is basically using an equation from the
notes. The circuit should look like this: (the order of
the components doesn’t matter)
1kΩ
6V
30pF
• When the switch is closed, the capacitor will start to
charge up and the voltage across it will increase like
this (you don’t need to draw the graph):
6V
4V
time
0V
τ=RxC
• It takes approximately one time constant, τ = R × C,
to rise to 4V (that is, 2/3 of the final voltage, which is
6V).
ELEC166 Worked Examples Week 4
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• Here, the time constant is 1kΩ × 30pF
= 103 × 30 × 10-12 = 30 × 10-9 sec, or 30 ns.
• The voltage across the capacitor and the resistor
must add up to 6V (by Kirchhoff’s Law), so that the
voltage across the resistor must be approximately
2V.
ELEC166 Worked Examples Week 4
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(Q4) A 100pF capacitor is connected in series with a
240V 50Hz AC voltage source. What is the AC
current flowing through the capacitor?
• The behaviour of a capacitor or an inductor alone in
an AC circuit is described by an equation that looks
and behaves very much like Ohm’s Law (which only
applies to resistors):
Resistor
V=I×R
Inductor or capacitor
v = i × XL
where
or
XC =
v = i × XC,
1
,
ωC
X L = ωL
• Note that this relates the magnitudes (amplitudes) of
the current and voltage, not their phases. Apart from
that, you can use it just like Ohm’s Law.
• Here, ω = 2π×50 rad/sec and C = 100 × 10-12 F, so
that
1
1
7
=
=
3
.
18
×
10
Ω
XC =
−12
ωC 2π × 50 × 100 × 10
• Hence
v
240V
i=
=
= 7. 5 µ A
7
X C 3.18 × 10 Ω
ELEC166 Worked Examples Week 4
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(Q14) Given the two voltage waveforms:
v1(t) = 6sin(200πt + 0.5)
v2(t) = 6sin(200πt)
(i)
(ii)
(iii)
(iv)
(v)
Calculate the period and the frequency.
Calculate the peak and the peak-to-peak
amplitudes.
Calculate the RMS amplitudes.
On the same graph, plot several cycles of v1(t) and
v2(t) vs time.
Comment on the plots.
• Notice that the two waveforms are identical sine
waves, apart from their phase. Hence the periods,
frequencies and amplitudes are the same.
• A sinusoidal waveform with the equation
v(t)=A sin (ωt + φ)
has angular frequency ω (=2π f) and peak amplitude A.
• If we write the two waveforms in the form
v1(t) = 6 sin(2π × 100 t + ….)
we can immediately see that the peak amplitude is
6 volts and that the frequency is 100 Hz.
ELEC166 Worked Examples Week 4
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• So the answers are:
(i) Frequency ( f ) = 100 Hz
Period = 1 / f = 1/100=0.01 sec.
(ii) The peak amplitude is 6 volts, and for a sine wave
the peak-to-peak amplitude is just twice the peak,
or 12 volts.
(iii) For a sine wave (only!) the RMS amplitude is just
equal to the peak divided by √2, or 6/1.4142 = 4.24
volts.
(iv) The graphs should look something like this:
8
6sin(200πt+0.5)
6
6sin(200πt)
4
2
Voltage (V)
0
-2
-4
-6
-8
0
0.005
0.01
0.015
0.02
0.025
0.03
Time (sec)
(v) Not much to say except what you probably have
already worked out by this stage: both plots will be
sine waves, with v1 leading v2 by 0.5 radians.
ELEC166 Worked Examples Week 4
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