PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
RLC CIRCUITS
Question 72
Consider a series RC circuit for which R = 2.0 × 106 Ω, C = 6.0 µF and E = 20 V. Find
(i) the time constant of the circuit,
(ii) the maximum charge on the capacitor after a switch in the circuit is closed, and
(iii) the current in the circuit at the instant just after the switch in the circuit is closed.
(i) The time constant τ is
τ = RC = 2.0 × 106 × 6.0 × 10−6 = 12 s
(ii) The maximum charge occurs when the capacitor has been charging for a long time, ie
as t → ∞:
t → ∞ ⇒ e−t/τ → 0
∴ q = EC(1 − e−t/τ )
⇒ EC
= 20 × 6.0 × 10−6
= 1.2 × 10−6 C
(iii) The maximum current I0 occurs at the moment just after the switch is closed, hence
at t = 0:
E
20
I0 =
=
= 10 µA
R
2.0 × 106
Question 73
A 2.0 nF capacitor with an initial charge of 5.1 µC is discharged through a 1300 Ω resistor.
Calculate
(i) the maximum current through the resistor,
(ii) the current through the resistor 9.0 µs after the resistor is connected across the terminals of the capacitor, and
(iii) the charge remaining on the capacitor after 8.0 µs.
(i) The charge on the capacitor is given by
q = EC(e−t/τ )
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
therefore the current at a time t is
d
q0 −t/τ
dq
=
EC(e−t/τ ) = −
(e
)
I=
dt
dt
RC
where q0 = EC is the maximum charge and τ = RC is the time constant. The
maximum current I0 occurs at time t = 0, therefore
I0 = −
q0
5.1 × 10−6
=−
= −1.96 A
RC
1300 × 2.0 × 10−9
(ii) At time t = 9.0 µs, the current is
I=−
q0 −t/τ
5.1 × 10−6
−6
−9
(e
)=−
(e−9.0×10 /(1300×2.0×10 ) ) = −6.61 × 10−2 A
−9
RC
1300 × 2.0 × 10
(iii) After 8.0 µs, the charge is
−6 )/(1300×2.0×10−9 )
q = q0 (e−t/τ ) = (5.1 × 10−6 ) × e−(8.0×10
= 2.35 × 10−7 C
Question 74
How many constants must lapse before a capacitor in a series RC circuit is charged to within
0.1% of its equilibrium charge?
To be within 0.1% of the equilibrium charge q0 , the charge on the capacitor must be
99.9
q0
q =
100
Sinceq = q0 (1 − e−t/τ )
99.9
⇒
q0 = q0 (1 − e−t/τ )
100
1
= e−t/τ
∴
1000
1
t
ln
= −
1000
τ
∴ t = τ ln 1000
= 6.9τ
It would take 6.9 time constants for the charge to be within 0.1% of the equilibrium charge
q0 .
Question 75
An electronic flash attachment for a camera produces a flash by using the energy stored in
a 750 µF capacitor. Between flashes, the capacitor recharges through a resistor whose resistance is chosen so that the capacitor will recharge with a time constant of 3.0 s. Determine
the value of the resistance.
τ = RC
∴R=
τ
3.0
=
= 4.0 × 103 Ω
−6
C
750 × 10
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 76
Three identical capacitors are connected with a resistor in two different ways. When they
are connected as in part (a) of the drawing, the time constant is 20 ms. What is the time
constant when they are connected with the same resistor as in part (b)?
(a)
The equivalent capacitance in part (a), CEQ , is given by
(a)
CEQ =
C ×C
3
+C = C
C +C
2
and in part (b) it is
(b)
CEQ =
2C × C
2
= C
2C + C
3
The time constant for (a) is then
3
(a)
τ (a) = RCEQ = RC = 20 ms
2
The time constant for (b) is
2
(b)
τ (b) = RCEQ = RC
3
Therefore the product R × C is
RC =
3 (b)
τ
2
3
RC = 20
2
3 3
∴ × τ (b) = 20
2 2
⇒ τ (b) = 8.9 s
and
Question 77
At what frequency does a 7.50 µF capacitor have a reactance of 168 Ω?
1
1
1
∴ω=
=
= 7.94 × 102 rad s−1
ωC
XC C
168 × 7.50 × 10−6
The cyclic frequency is then
XC =
f=
ω
7.94 × 102
=
= 126 Hz
2π
2π
Question 78
What is the inductance of an inductor that has a reactance of 1.8 kΩ at a frequency of
4.2 kHz?
XL = ωL
∴L=
1.8 × 103
XL
=
= 6.8 × 10−2 H
ω
2 × π × 4.2 × 103
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 79
What voltage is needed to create a current of 29.0 mA in a circuit containing only a 0.565 µF
capacitor, when the frequency is 2.60 kHz?
The maximum voltage V0 is given by
V 0 = I 0 XC =
I0
29.0 × 10−3
=
= 3.14 V
ωC
2 × π × (2.60 × 103 ) × (0.565 × 10−6 )
Question 80
Three capacitors are connected in parallel across the terminals of a 440 Hz generator. The
capacitances are 2.00, 4.00 and 7.00 µF. (a) Find the equivalent capacitance of these capacitors. (b) If the generator supplies a total current of 0.620 A, what is the voltage of the
generator? (c) What is the current supplied to each of the three capacitors?
(a) The equivalent capacitance CEQ is
CEQ = 2.00 + 4.00 + 7.00 = 13.0 µF
(b) The maximum voltage is
V 0 = I 0 XC =
0.62
I0
=
= 17.3 V
ωC
2 × π × 440 × 13 × 10−6
(c)
V 0 = I 0 XC
∴ I0 = ωCV0
= 2πf V0
= 2 × π × 440 × 2.00 × 10−6
= 9.54 × 10−2 A for the 2.00 µF capacitor
= 1.91 × 10−1 A for the 4.00 µF capacitor
= 0.333 A for the 7.00 µF capacitor
Question 81
A coil has a resistance of 20 Ω. At a frequency of 100 Hz, the voltage across the coil leads
the current in it by 30°. Determine the inductance of the coil.
The voltage is given by
V = V0 cos(ωt + α)
where the maximum voltage V0 is
V0 = I 0 X = I 0
q
R2 + XL2
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
and the phase angle α is
(L)
tan α =
V0
(R)
V0
=
ωL
XL
=
R
R
The inductance is therefore
L=
20
R
tan α =
tan 30° = 84 × 10−2 H
ω
2π100
Question 82
A series circuit has an impedance of 50 Ω and a power factor of 0.6 at 60 Hz, the voltage
lagging the current.
(a) Should an inductor or capacitor be placed in series with the circuit to raise its power
factor?
(b) What size element will raise the power factor to unity?
(a) The voltage lags the current so the circuit is capacitative, hence an inductor in series
will increase the power factor (cos α).
(b) A power factor cos α = 0.6 gives a phase angle α = −53.13°. The impedance Z is 50 Ω,
hence
p
Z =
R2 + (XL − XC )2 = 50
∴ R2 + (XL − XC )2 = 2500
2
2500
XL − XC
=
−1
R
R2
= tan2 α
1
=
−1
cos2 α
1
=
−1
0.62
100 36
=
−
36
36
64
=
36
16
=
9
2500
16
∴
−1 =
R2
9
25
2500
=
R2
9
∴ R = 30 Ω
The power factor is required to be unity, hence cos α = 1 ⇒ α = 0. Let XL′ be the
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
added reactance, such that XL′ + XL XC = 0. But from above
2
XL − XC
2500
16
=
=
2
R
R
9
4
XL − XC
= ±
∴
R
3
4
XL′ =
R
3
since X cannot be negative. The added reactance must be
4
ωL = 30 = 40 Ω
3
⇒L=
40
= 0.106 H
2π × 60
Question 83
A series circuit has a resistance of 75 Ω and an impedance of 150 Ω. What power is consumed
in the circuit when 120 V (rms) is impressed across it?
The impedance and the resistance are related by
Z 2 = R2 + (XL − XC )2
with the phase angle α
tan α =
XL − XC
R
√
Z 2 − R2
√ R
1502 − 752
=
√ 75
3
=
∴ α = 60°
=
The average power hP i is
hP i = Vrms Irms cos α =
2
Vrms
1202
cos 60° =
cos 60° = 48 W
Z
150
Alternatively, since Z = 150 Ω and R = 75 Ω, from a phasor diagram
cos α =
75
R
=
Z
150
So the average power is
hP i =
2
Vrms
1202
cos α =
× f rac75150 = 48 W
Z
150
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 84
A resistance of 500 Ω and capacitance of 2.00 µF are connected in series across an AC source
(C)
with emf amplitude of 282 V and frequency 60 Hz. Find XC , Z, I0 , V0 and α.
Reactance:
XC =
Impedance
Z=
Maximum voltage is
1
1
=
= 1.33 × 103 Ω
−6
ωC
2 × π × 60 × 2.0 × 10
q
p
R2 + XC2 = 5002 + (1.33 × 103 )2 = 1.42 × 103 Ω
V0 = I 0 Z
∴ I0 = VZ0
=
=
282
1.42 × 103
0.199 A
The maximum voltage across the capacitor is then
(C)
V0
= I0 XC = 0.199 × 1.33 × 103 = 2.64 × 102 V
The phase angle
1.33 × 103
XC
=
⇒ α = −69.8°
R
500
The angle is negative, since the voltage lags the current for a capacitor.
tan α =
Question 85
An inductor having a reactance of 25 Ω and a resistance R gives off heat at the rate of 10 W
when it carries a current of 0.50 A (rms). What is the impedance of the inductor?
The average power dissipated through the resistance R is
2
hP i = Vrms Irms = Irms
R
therefore the resistance R is given by
R=
The impedance is then
10
hP i
=
= 40 Ω
2
Irms
0.52
q
√
Z = R2 + XL2 = 402 + 252 47 Ω
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 87
In a series RLC circuit the instantaneous current is
I = 4.00 cos 1200t A
where t is in seconds. Given that R = 40.0 Ω, L = 40.0 mH and C = 25.0 µF, find the
instantaneous across R, L and C and also across L and C in series.
From the above equation, I0 = 4.00 A and ω = 1200 rad s−1 . The impedance Z is
p
Z =
R2 + (XL − XC )2
s
2
1
R2 + ωL −
=
ωC
s
2
1
402 + 1200 × 40 × 10−3 −
=
1200 × 25 × 10−6
= 42.6 Ω
The maximum voltage V0 is given by
V0 = I0 Z = 4.00 × 42.6170 V
The phase angle is
α = tan
−1
XL − XC
R
= 20.1°
Here, α is positive, hence XL > XC . The instantaneous voltage is then
V = 170 cos(1200t + 0.351) V
where α = 0.351 rad.
For the L and C combination, the impedance is
Z = |XL − XC | = 14.7 Ω
The maximum voltage across the combination is
(LC)
V0
= I0 Z = 4.00 × 14.7 = 58.8 V
Since XL > XC , the voltage across the LC combination leades the current by 90°. The
instantaneous voltage is
(LC)
V (LC) = V0
cos(ωt + α) = 58.8 cos(1200t + π/2) V
Question 89
In a series RLC circuit the voltage amplitude is 50.0 V and the angular frequency is 1000 rad s−1 .
Let R = 300 Ω, L = 0.900 H and C = 2.00 µF. Calculate
(a) the impedance of the circuit,
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(b) the current amplitude,
(c) the voltage amplitude across the resistor and across the inductor,
(d) the phase angle between V (t) and I(t).
(a) Impedance
p
Z = R2 + (XL − XC )2 =
s
3002
1
+ 1000 × 0.900 −
1000 × 2.00 × 10−6
I0 =
V0
50.0
=
= 0.100 A
Z
500
2
= 500 Ω
(b) Maximum current
(c) Maximum voltage across R
(R)
V0
= I0 R = 0.100 × 300 = 30.0 V
Maximum voltage across L
(L)
V0
= I0 XL = I0 ωL = 0.100 × 1000 × 0.900 = 90.0 V
(d) The phase angle α
XL − XC
1000 × 0.900 − 1/(1000 × 2.00 × 10−6 )
−1
−1
α = tan
= tan
= 53.1°
R
300
Question 90
The power dissipated in a series RLC circuit is 65.0 W and the current is 0.530 A. The circuit
is at resonance. Determine the peak voltage of the generator.
The average power hP i is
hP i = Vrms Irms cos α
At resonance α = 0, hence
V0 =
√
2 Vrms =
√ hP i √ 65.0
2
= 173 V
= 2
Irms
0.530
Question 91
At resonance in a series RLC circuit, the rms current is 0.100 A and the rms potential differences across R and C are 40.0 V and 20.0 V respectively. Find the rms potential difference
across L. Calculate the impedance in the circuit at twice the resonant frequency.
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PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
At resonance, V (L) = V (C) = 20.0 V.
The resisntance, which is independent of frequency is
R=
40.0
Vrms
=
= 400 Ω
Irms
0.100
At twice the resonant frequency, the reactances are
XL = 2ω0 L
(L)
= 2X0
20.0
= 2
0.100
= 400 Ω
1
XC =
2ω0 C
1 (C)
=
X
2 0
1 20.0
=
2 0.100
= 100 Ω
p
∴Z =
R2 + (XL − XC )2
p
400+ (400 − 100)2
=
= 500 Ω
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