ECSE 362
Winter 2023
McGill University – ECSE 362
Tutorial Assignment Solutions – Week 1
1. Consider the oscillograms in Figure 1. They display the ac voltage across an electrical
load and the ac current flowing through it.
Figure 1: Load ac voltage v(t) and current i(t).
From the oscillograms, determine:
(a) The frequency of the voltage and current in hertz.
Answer: By inspection of the oscillograms, we find that one period T of either
the current or the voltage lasts approximately 0.0167 s. Thus, the frequency is
f ≈ 1/T = 60 Hz.
(b) The rms voltage and current values in volts and amperes, respectively.
Answer: By inspection of the oscillograms, we find that the voltage waveform √
has a
peak amplitude of approximately 170 V. Therefore, the rms voltage V = 170/ 2 ≈
120 V. Similarly, we find that the
√ peak current is equal to about 10 A. This means
that the rms current is I = 10/ 2 ≈ 7.07 A.
(c) The value of the phase difference between the voltage and the current. Take the
voltage sinusoid as the reference. State your answer in degrees.
Answer: Taking the voltage sinusoid as the reference, we find that the first zero
crossing of the voltage occurs at t ≈ 0.0042 s. For the current, we find that the
first zero crossing occurs at t ≈ 0.0014 s. From this information, we infer that
there is a delay ∆t = 0.0014 − 0.0042 = −0.0028 s between the current and the
voltage waveforms (i.e., the current leads the voltage here) . In angular terms this
is equivalent to ϕ = +2π · ∆t/T = 2π · 0.0028/0.0167 ≈ 1.05 rad ≈ 60.4◦ .
© François Bouffard, 2023, all rights reserved.
1
ECSE 362
Winter 2023
Note: This can be done using any other reference point on the respective sinusoids
(e.g., peaks).
(d) The phasor representations of the voltage and of the current.
√
Answer: Our sinusoid/phasor convention states that u(t) = 2U cos(ωt + θ) ⇔
U = U ejθ . Given the information found above, we have:
◦
V = 120ej0 = 120 0◦ V
◦
I = 7.07ej60 = 7.07 60◦ A
(e) The impedance of the load in ohms. Is this an R-L or an R-C load? Justify your
answer.
Answer: By definition
Z=
V
120 0◦
=
= 17.0 −60◦ Ω
I
7.07 60◦
This is an R-C load since the current is leading the voltage (or, equivalently since
the angle of the impedance is negative-valued).
(f) How much active and reactive power is consumed by this load?
Answer: We have the complex power S = VI∗ = 120 0◦ · 7.07 −60◦ = 848 −60◦ .
Thus, P = S cos ϕ = 424 W and Q = S sin ϕ = −734 var. Here the load is producing
reactive power.
2. The circuit diagram in Figure 2 represents a single-phase three-wire household distribution
system. It consists of two series-connected 120 V, 60 Hz ac sources which permit the
supply of both 120 V and 240 V household loads.
I1
+
V1 = 120 0◦ V
I2
Zx = 8 0◦ Ω
Zz = 11 15◦ Ω
+
V2 = 120 0◦ V
I3
Zy = 16 36.9◦ Ω
Figure 2: Single-phase, three-wire household distribution system.
(a) Determine the current flowing through each load.
© François Bouffard, 2023, all rights reserved.
2
ECSE 362
Winter 2023
Answer: The current through Zx is Ix = V1 /Zx = 120 0◦ /8 0◦ = 15 0◦ A.
The current through Zy is Iy = V2 /Zy = 120 0◦ /16 36.9◦ = 7.5 −36.9◦ A.
Finally, the current in Zz is Iz = (V1 + V2 )/Zz = 240 0◦ /11 15◦ = 21.8 −15◦ A.
(b) Determine the currents I1 , I2 and I3 .
Answer: By KCL, I1 = Ix + Iz = 36.1 − j5.6 = 36.5 −8.9◦ A; I2 = −Ix + Iy =
−9.0 − j4.5 = 10.1 −153.4◦ A; and I3 = −Iy − Iz = −27.1 + j10.1 = 28.9 159.4◦ A.
(c) Determine the active and reactive power consumed by each load.
Answer: We have Sx = Vx I∗x = 120 0◦ · 15 0◦ = 1800 + j0 VA, i.e., Px = 1.80 kW
and Qx = 0.00 var.
Similarly, Sy = Vy I∗y = 120 0◦ · 7.5 36.9◦ = 720 + j540 VA, i.e., Py = 720 W and
Qy = 540 var.
Finally, Sz = Vz I∗z = 240 0◦ · 21.8 15◦ = 5054 + j1354 VA, i.e., Pz = 5.05 kW and
Qz = 1.35 kvar.
(d) Determine the active and reactive power supplied by each source.
Answer: S1 = V1 I∗1 = 120 0◦ · 36.5 8.9◦ = 4380 8.9◦ = 4327 + j678 VA, i.e.,
P1 = 4.33 kW and Q1 = 678 var.
Similarly, S2 = V2 (−I∗3 ) = 120 0◦ · 28.9 20.6◦ = 3468 20.6◦ = 3246 + j1220 VA, i.e.,
P2 = 3.25 kW and Q2 = 1.22 var.
We note that Px + Py + Pz = P1 + P2 = 7.57 kW and Qx + Qy + Qz = Q1 + Q2 =
1.89 kvar, as it should.
(e) Argue why high-power loads such as cookers and clothes dryers are fed using 240 V
rather than 120 V.
Answer: High power domestic loads (typically above 1.3-1.5 kW, i.e., your typical
clothes iron) would draw too much current at 120 V. Thus, for safety reasons highpower appliances are fed with 240 V. We see that for the same power, doubling the
voltage allows for a reduction of current magnitude by a factor of two, and at the
same time wiring heating by a factor of four.
© François Bouffard, 2023, all rights reserved.
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