Q1. (a) Prove from first principles that two capacitors C1 and C2 connected in series can be replaced
with an equivalent capacitance of C1C2 / (C1+C2). [2 marks]
4 F
3 F
R1
R2
C
100V
4 F
6 F
100V
R2
Figure Q1-(b)
R1
Figure Q1-(c)
(b) Find the voltage across each capacitor of the circuit in Figure Q1-(b). [4 marks]
(c) Find an expression for the charging time constant of the capacitor C shown in Figure Q1-(c).
Also find the steady state voltage across the capacitor. [4 marks]
ANSWER
(a) Let Q1 be the charge on C1 and Q2 charge on C2,
then Q1 = C1V1 and Q 2 = C2V2 .
For the equivalent capacitor C, we have Q = CV
Since the two capacitors are in series,
Q1 = Q 2 = Q , Hence we have,
⎡1
1 ⎤
Q Q
+
= Q ⋅ ⎢ + ⎥,
C1 C 2
⎣ C1 C 2 ⎦
Comparing with Q = CV , we get
V = V1 + V2 =
1
1
1
=
+
or
C C1 C 2
C=
C1C 2
C1 + C 2
(b) The equivalent circuits are given below,
Using the voltage division rule, we get the following
–1–
(c) Get the Thevenin’s equivalent across the capacitor.
Rth = (R1||R2) + (R1||R2), therefore, Rth =2. R1.R2/(R1+R2)
Also,
⎛ R2
⎛ R − R1 ⎞
R1 ⎞
⎟⎟ = E ⎜⎜ 2
⎟⎟
−
Eth = E ⎜⎜
⎝ R1 + R2 R1 + R2 ⎠
⎝ R1 + R2 ⎠
Hence,
The time constant = Rth. C = 2. R1.R2.C/(R1+R2)
⎛ R − R1 ⎞
⎟⎟
Final Voltage = Eth = 100.⎜⎜ 2
R
R
+
1
2
⎠
⎝
Q2. (a) Write a differential equation relating output current i(t) to the input voltage v(t) for the
circuit in Figure Q2. [5 marks]
(b) By means of the differential equation derived above, answer the following.
i. What is the order of the differential equation? How is the order of the differential
equation related to the circuit in question? [1 mark]
ii. If the input voltage v(t) is a DC source of magnitude E, what is the current at steady
state? [1 mark]
iii. If the input voltage v(t) is an AC source of angular frequency ω, derive an expression
for the input impedance of the circuit. [2 marks]
iv. Validate your answer in part (iii) above using AC theory. [1 mark]
ANSWER
(a) We can write the following equations for V1, V2 and V.
V = V1 + V2 ………………… (1)
i = i1 +i 2 …………………… (2)
di
V1 = L1 2 ………………… (3)
dt
1
V1 =
i1 dt ………………. (4)
C1 ∫
di
V2 = iR2 + L2 …………… (5)
dt
di 1
d 2V
= C1 21 , and combining with equation (3),
dt
dt
2
V
d V1 V1
1 di d 2V1
di di1 di2
= 2 + 1 ………… (6)
=
+
= C1 2 + , or
we get,
dt dt dt
L1
C1 dt
dt
L1C1
dt
Differentiating equation (4) twice yields
d 2i
d 3 i d 2V 2
Also, differentiating equation (5) twice yields R 2 2 + L2 3 =
…… (7)
dt
dt
dt 2
Adding (6) and (7),
V
d 2V2
d 2i
d 3i d 2V
1 di
+ R2 2 + L2 3 = 21 + 1 +
…………
C1 dt
L1C1 dt 2
dt
dt
dt
–2–
(8)
Substituting from equation (1) and its derivatives, (8) becomes
1 di
1
d 2i
d 3i d 2V
(V − V2 )
+ R2 2 + L2 3 = 2 +
C1 dt
dt
dt
dt
L1C1
=
1 ⎛
d 2V
di ⎞
+
⎜V − iR2 − L2 ⎟
2
dt
L1C1 ⎝
dt ⎠
Re-arranging the terms,
R2
d 3i R2 d 2 i 1 ⎛ 1
1 ⎞ di
1 d 2V
1
⎜
⎟
+
⋅
+
+
+
i
=
⋅ 2 +
⋅V
3
2
⎜
⎟
dt
L2 dt
C1 ⎝ L1 L2 ⎠ dt L1 L2 C1
L2 dt
L1 L2 C1
(b) Using the above differential equation,
i. The order of the differential equation is 3. This is because the circuit has 3 independent
energy storing elements.
ii. To find steady state current, substitute di/dt= 0 and dV/dt=0 (all higher derivatives are
also zero). Thus, I = E/R2.
iii. Substitute d/dt = jω, (and swapping left and right hand sides)
⎡
⎡1
R2
R2 ⎤
1 ⎤
1 ⎛1
1 ⎞
2
2
3
⋅ ( jω ) + ⎜⎜ + ⎟⎟( jω ) +
⎥⋅I
⎢ ⋅ ( jω ) +
⎥ ⋅ V = ⎢ ( jω ) +
L1 L2 C1 ⎦
L2
C1 ⎝ L1 L2 ⎠
L1 L2 C1 ⎦
⎣ L2
⎣
⎡
⎡ 1 2
R2
R2 ⎤
1 ⎤
1 ⎛1
1 ⎞
3
⋅ ω 2 + j ⎜⎜ + ⎟⎟ω +
⎥⋅I
⎢− ω +
⎥ ⋅ V = ⎢ − jω −
L1 L2 C1 ⎦
L
C
L
L
L
L
C
⎣ L2
2
1
1
2
1
2
1
⎝
⎠
⎣
⎦
2
3
2
1 − ω L1C1
− jω L1 L2 C1 − R2 L1C1ω + j (L1 + L2 )ω + R2
⋅V =
⋅I
L1 L2 C1
L1 L2 C1
(
)
[
]
(
)
(
(
V R2 1 − ω 2 L1C1 + jω L1 + L2 − ω 2 L1 L2 C1
=
I
1 − ω 2 L1C1
)
jωL1
1 − ω 2 L1C1
= R 2 + jω L 2 +
(
)
iv. Using AC theory, the impedance is
1
jωC1
Z = R2 + jωL2 +
1
jωL1 +
jωC1
jωL1 ⋅
= R2 + jωL2 +
jωL1
1 − ω 2 L1C1
(
–3–
)
)
Q3. (a) Derive an expression for the total energy stored in the
coupled circuit in Figure Q3-(a). [2 marks]
(b) The circuit in Figure Q3-(b) is supplied by a sinusoidal
voltage source Vs of variable frequency. The output Vout
is taken across the inductor.
i. Write an expression for the output voltage Vout as a
function of the input voltage Vs . [1 mark]
ii. Find an expression for the resonance frequency of the
circuit. [1]
iii. Show that half-power frequencies are given by, [2 marks]
R
Vs
C
L
Vout
1
1
1
1
+
+
Figure Q3-(b)
2
LC
4πRC 2π (2 RC )
iv. If the circuit is to be designed with a resonance frequency of 3.2 kHz and a bandwidth
of 1 kHz, calculate the appropriate values for L and C, given that R = 2.0 kΩ. [2 marks]
v. With the circuit parameters as found in part (iv) above, determine the output voltage at
the frequencies of f1 and f2, when Vs = 10V. [2 marks]
f1 , f 2 = ±
ANSWER
(a) Total energy stored
W = ∫ v1i1dt + ∫ v2 i2 dt
di ⎞
di ⎞
⎛ di
⎛ di
= ∫ ⎜ L1 1 + M 2 ⎟i1dt + ∫ ⎜ L2 2 + M 1 ⎟ i2 dt
dt ⎠
dt ⎠
⎝ dt
⎝ dt
= ∫ L1i1di1 + ∫ L2 i2 di2 + M ∫ (i1di2 + i2 di1 )
= 12 L1i 21 + 12 L2 i 22 + Mi1i2
(b) (i). The output voltage is given by,
Z
Vout =
⋅ Vs , where Z is the impedance of L and C in parallel.
R+Z
jωL
=
(1-ω LC ) ⋅ V
s
jωL
R + (1-ω LC )
2
2
⎛
⎞
jωL
⎟⎟ ⋅ Vs
= ⎜⎜
2
⎝ R 1 − ω LC + jωL ⎠
⎡
⎤
⎢
⎥
1
=⎢
⎥ ⋅ Vs
R
2
⎢1 − j
1 − ω LC ⎥
ωL
⎣
⎦
(ii). The resonance frequencies are given by unity power factor,
R
1 − ω 2 LC = 0
ωL
1
ω0 =
LC
(iii). At half-power frequencies,
(
)
(
(
)
)
–4–
⎡
⎤
⎢
⎥
V
1
⋅ Vs = s , Since at resonance the output voltage is Vs
⎢
⎥
2
⎢1 − j R 1 − ω 2 LC ⎥
ωL
⎣
⎦
R
i.e.,
1 − ω 2 LC = ±1
ωL
Taking the positive sign,
(
)
(
)
R − ω 2 RLC = ωL
⎛ 1 ⎞
⎛ 1 ⎞
⎟ω − ⎜
⎟=0
⎝ RC ⎠
⎝ LC ⎠
ω2 + ⎜
ω1 = −
1
+
2 RC
1
1
+
, Note that the other solution yields a negative
2
(2 RC ) LC
frequency and hence it is ignored.
Taking the nagative sign,
⎛ 1 ⎞
⎛ 1 ⎞
⎟ω − ⎜
⎟=0
⎝ RC ⎠
⎝ LC ⎠
ω2 −⎜
ω2 =
1
+
2 RC
1
1
+
, Note that the other solution yields a negative
2
(2 RC ) LC
frequency and hence it is ignored.
1
1
1
1
+
+
2
4πRC 2π (2 RC ) LC
(iv). Resonnance frequency,
1
1
f 0=
= 3200 Hz
2π LC
1
LC =
= 2.474 × 10 −9
4π 2 f 02
Bandwidth is given by,
1
f1 − f 2 = 2 ×
= 1000 Hz
4πRC
1
RC =
= 1.592 × 10 − 4
2π (BW)
f1 , f 2 = ±
Thefore, C = 0.08 µF and L = 31 mH.
(v). The output voltage at half-power frequencies must be 1/√2 of peak output at resonance.
Since the peak output is 10V, the output at half-power frequencies is 7.07V.
–5–
Q4. (a)
(b)
(c)
(d)
For the circuit in Figure Q4, find the input impedance across XY. [2 marks]
Find the short circuit current through XY. [2 marks]
Find the Norton’s equivalent circuit across XY. [2 marks]
If the voltage source is moved to the terminals XY, What
would be the Norton’s equivalent circuit across AB? [2 marks]
(e) Are the two Norton’s equivalent circuits identical? Comment
on your results. [2 marks]
ANSWER
(a) The impedance across XY, is ZXY .
Z1 = 8 j15 = 6.2284 + j 3.3218
Z 2 = Z1 − j 40 = 6.2284 -j36.6782
Z 3 = Z 2 50 = 18.8100 -j 20.3455
Z XY = Z 3 − j 20 = 18.8100 -j 40.3455
1
= 0.0095 + j 0.0204
Z XY
(b) First, find the total impedance when XY are short circuited.
Z 4 = − j 20 50 = 6.8966 - j17.2414
YXY =
Z 5 = Z 4 − j 40 = 6.8966 - j 57.2414
Z 6 = Z 5 j15 =0.8471 + j 20.1882
Z sc = Z 6 + 8 = 8.8471 + j 20.1882
20
= 0.3642 - j 0.8311, this is the s/c current at the source.
8.8471 + j 20.1882
Using current division rule successively,
j15
I2 =
I 1 = -0.0790 + j 0.3080
Z 5 + j15
∴ I1 =
50
I 2 = -0.1744 + j 0.2383
50 − j 20
(c) Hence, the Norton equivalent circuit has a current source of (-0.1744 + j0.2383) A and a
parallel admittance of YXY = 0.0095 + j0.0204
(d) When the voltage source is moved to XY, using the reciprocity theorem, the short circuit
current across AB must be the same as the short circuit current Isc found earlier.
By inspection, the input impedance across AB, must be the same as Zsc calculated in part (b)
above.
(e) The two equivalent circuits are not identical. However, they have the same current source
but different admittances in parallel.
I sc =
–6–
Q5. (a) An alternating voltage source Vs has an internal impedance of r + jx. Find an expression for
the purely resistive load Rp which when connected to the source, maximises the real output
power. [4 marks]
(b) Derive an expression for the source efficiency for the above circuit when it delivers
maximum power. [3 marks]
(c) You need to design a 10V, AC source such that for purely resistive loads, it delivers a
maximum of 12.5W at a source efficiency of 75 percent. Determine the internal impedance
of the source. [3 marks]
ANSWER
(a) When a purely resistive load R is connected, the power output is given by
2
V
P=I R=
⋅R
r + jx + R
2
V2
⋅R
(R + r )2 + x 2
For maximum power, dP/dR = 0
=
[
]
dP
2
= V 2 ⋅ R (2 ( r + R ) ) − V 2 ⋅ (r + R ) + x 2 ⋅ 1 = 0
dR
2
2 R (r + R ) = (r + R ) + x 2
R2 = r 2 + x2
∴ Rp =
r 2 + x2
(b) The source efficiency at maximum power,
2
Im Rp
Output Power
r 2 + x2
=
=
η=
Total Power at source I m 2 (R p + r ) r + r 2 + x 2
( r)
1 + (x )
r
2
1+ x
=
2
1+
(c) The maximum power delivered to the load,
2
PL = I m R p
=
=
(r +
V2
r +x
2
2
) +x
2
⋅ r 2 + x2
2
V2
2(r + x ) + 2r r + x
2
2
2
( )
2
⋅ r 2 + x2
2
⎡
⎤
x
1
+
V ⎢
⎥
r
=
⋅⎢
⎥
2
2
2r ⎢
⎥
1+ x
+ 1+ x
r
r ⎦
⎣
From the equation for efficiency, η = 0.75, we get (x/r)2 = 8 or x/r = 2.828.
By substituting for x/r in equation for PL, we get, r = 1.0Ω. Hence, x = 2.828Ω.
2
( )
( )
–7–
Q6. (a) Find the ABCD parameters for the transformer circuit in Figure Q6-(a). [5 marks]
Z1
Z2
Z3
Figure Q6-(b)
(b) Find appropriate values for Z1, Z2 and Z3 of the T-network in Figure Q6-(b) such that it is
equivalent to the transformer circuit in Figure Q6-(a). [5 marks]
ANSWER
(a) Writing the Kirchhoff’s law for the two loops,
V1 = I1 R1 + jωL1 I1 − jωMI 2
V2 = I 2 R2 − jωL2 I 2 + jωMI1
Using the above equations together with the definitions for A, B, C and D,
A=
B=
V1
V2
V1
V2
=
R1 I1 + jωL1 I1 R1 + jωL1
=
jωMI1
jω M
=
I
I1 R1 + jωL1 I1 − jωMI 2
= (R1 + jωL1 ) 1 − jωM
I2
I2
I 2 =0
V2 = 0
But from second equation above,
∴ B = (R1 + jωL1 )
C=
D=
I1
V2
I1
I2
=
I 2 =0
=
V2 = 0
I1 (R2 + jωL2 )
=
I2
jωM
(R2 + jωL2 ) − jωM =
jω M
[
1
(R1 + jωL1 )(R2 + jωL2 ) − ω 2 M 2
jω M
I1
1
=
jωMI1 jωM
(R2 + jωL2 )
jω M
(b) For the T-network,
⎡ A B ⎤ ⎡1 Z1 ⎤ ⎡ 1 0⎤ ⎡1 Z 2 ⎤
⎢C D ⎥ = ⎢0 1 ⎥ ⎢Y 1⎥ ⎢0 1 ⎥
⎦⎣ 3
⎦
⎣
⎦ ⎣
⎦⎣
⎡(1 + Z1Y3 ) Z 2 + (1 + Z 2Y3 )Z1 ⎤
=⎢
(1 + Z 2Y3 ) ⎥⎦
⎣ Y3
By comparing the two sets of parameters,
–8–
]
For parameter C,
1
which yields Z 3 = jωM
jω M
For parameter A,
Y3 =
(1 + Z1Y3 ) = R1 + jωL1
jω M
Substituting for Y3 , Z1 = R1 + jωL1 − jωM = R1 + jω (L1 − M )
Similarly for parameter C,
(1 + Z 2Y3 ) = R2 + jωL2
jωM
Substituting for Y3 , Z 2 = R2 + jωL2 − jωM = R2 + jω (L2 − M )
Q7. (a) Using the first principles and general properties of Laplace transform, find the Laplace
transform of causal time function f (t ) = e − at sin ωt . [2 marks]
(b) In the circuit of Figure Q7, the capacitor has an initial charge of 500µC. If the switch S is
closed at t = 0, using Laplace transform method, find an expression in time t for the
subsequent variation of current i(t). [5 marks]
(c) If L=1 mH, R1=400Ω, R2=600Ω and C=25µF, determine the following values related to i(t),
[2 marks]
iv. initial current,
v. time at which current becomes zero momentarily,
vi. peak value of current and the time it occurs,
vii. steady state value of current.
(d) Sketch, approximately to scale, the current wave form i(t) from t = 0 to 100ms. [1 mark]
ANSWER
(a) See class notes for the Laplace transform of sin ωt . Multiplying by e-at in time domain will
have the effect of shifting sÆ(s+a) in frequency domain. Combining these two results, we
get F ( s ) =
ω
(s + a )2 + ω 2
.
(b) When converted to frequency domain the circuit becomes (the initial charge is equal to 20V
step input in series with the capacitor),
I2(s)
100/s
Ls
V2
V1
–9–
R2
I(s)
R1
1/Cs
20/s
And applying Kirchhoff’s law’s,
1
20
V1 = R2 I ( s ) +
I ( s) +
……………………
Cs
s
100
= V1 + V2
…………………………
s
(1)
(2)
Equation (2) yields,
100
= V1 + Ls ⋅ I 2 ( s )
s
⎛
V ⎞
= V1 + Ls ⋅ ⎜⎜ I ( s ) + 1 ⎟⎟
R1 ⎠
⎝
⎛ Ls ⎞
= V1 ⎜⎜1 + ⎟⎟ + Ls ⋅ I ( s )
⎝ R1 ⎠
Substituting for V1 from (1),
1
20 ⎞ ⎛ Ls ⎞
⎛
I ( s ) + ⎟ ⋅ ⎜⎜1 + ⎟⎟ + Ls ⋅ I ( s )
= ⎜ R2 I ( s ) +
Cs
s ⎠ ⎝ R1 ⎠
⎝
⎡⎛
⎤
1 ⎞⎛ Ls ⎞
20 20 L
+
= ⎢⎜ R2 +
⎟⎜⎜1 + ⎟⎟ + Ls ⎥ ⋅ I ( s ) +
Cs
R
s
R1
⎝
⎠
1
⎝
⎠
⎣
⎦
Solving for I(s),
⎛
⎞
80 − 20⎜ L ⎟ s
R
⎝
1⎠
I (s ) =
⎛ R + R2 ⎞ 2 ⎛ L + R1 R2 C ⎞
1
⎟⎟ s + ⎜⎜
⎟⎟ s +
L⎜⎜ 1
C
⎝ R1 ⎠
⎝ R1C ⎠
The above expression has a second order denominator and a first order numerator.
In general this expression can be factored as,
a −b
I (s ) = k ⋅
(s + a )(s + b )
Hence the time domain expression for current i(t) is,
i (t ) = k1 ⋅ e − at − k 2 ⋅ e − bt
(c) Substituting values,
80 − 5 × 10 −5 s
I (s ) =
0.0025 s 2 + 600 s + 40000
3.2 × 10 4 − 0.02s
= 2
s + 2.4 × 10 5 s + 1.6 × 10 7
Using partial fraction expansion, we get
0.1334
0.1534
I (s) =
−
(s + 66.7 ) s + 2.4 ×105
In time domain,
5
i (t ) = 0.1334 ⋅ e −66.7t − 0.1534 ⋅ e −2.4×10 t
(
)
–10–
The initial current: substitute t=0;
i(0) = 0.1334 – 0.1534 = 20 mA.
The at which current crosses the zero axis:
5
0.1334 ⋅ e −66.7 t − 0.1534 ⋅ e −2.4×10 t = 0
0.1334 ⋅ e −66.7 t = 0.1534 ⋅ e − 2.4×10
5
t
Taking log on both sides,
ln (0.1334 ) − 66.7t = ln (0.1534 ) − 2.4 × 105 t
Thus t0 = 0.58×10-6 sec.
Peak value of current:
5
di
= 0.1334 × (− 66.7 ) ⋅ e −66.7t − 0.1534 × − 2.4 × 10 5 ⋅ e −2.4×10 t = 0
dt
(
8.89778 × e −66.7t = 36816 × e − 2.4×10 t
5
⎛ 36816 ⎞
5
− 66.7t = ln⎜
⎟ − 2.4 × 10 t
⎝ 8.89778 ⎠
5
2.3993 × 10 t = 8.3279
t = 34.7 × 10 −6
Corresponding current would be
−6
5
−6
i (t ) = 0.1334 ⋅ e −66.7×34.7×10 − 0.1534 ⋅ e −2.4×10 ×34.7×10
= 133 mA
The steady state value of current is zero.
–11–
)