Question #1 (30 Marks):
For the circuit shown in Fig. 1, find out:
(a) the Thevenin equivalent circuit for the network external to the resistor R.
(b) the value of R for maximum power to R.
(c) the maximum power to R.
Fig. 1
a)
5
RTH ¼
+
I 25Ω =
I25Ω
ETH
+
− V3A
−
5
RTH = 25 || (10 + 6) =
I=3A
( 25) (16)
= 9.756 Ω 5
25 + 16
16
(3 A) = 1.171 A ,
25 + 16
V3 A = I 25Ω ( 25) = (1.171)(25) = 29.275 V
ETH = V3 A − 20 = 29.275 − 20 = 9.275 V 5
b) R = RTH = 9.756 Ω 3
c) PMAX
2
ETH
(9.275) 2
=
=
= 2.2044 W 4
4 RTH ( 4)(9.756)
3
Question #2 (30 Marks)
For the circuit shown in Fig. 2, find the Norton equivalent circuit for the network external
to the resistor R. Also, using the equivalent circuit, find the current passing through the
resistor R if R = 6 Ω.
Fig. 2
5
5
» RN
IN
RN = (( R1 || R2 ) + R3 ) = ((5 || 5) + 5)
= 2.5 + 5 = 7.5 Ω 5
R2
E
(
)
IN =
R1 + ( R2 || R3 ) R2 + R3
5
I
20
5
(
)
5 + (5 || 5) 5 + 5
20 ⎛ 1 ⎞
=
⎜ ⎟ = 1.333 A 5
7.5 ⎝ 2 ⎠
=
I=
RN
7.5
(1.333) = 0.741 A 5
IN =
RN + R
7.5 + 6
Page (2 of 4)
EE308 Second Midterm Exam 1429/1430
Sunday 28/1/1430H
Question #3 (20 Marks)
For the following pairs of voltages and currents, indicate whether the element involved is a
capacitor, an inductor, or a resistor, and determine the value of C, L, or R. Also, determine
the power absorbed in watts for each set:
a) v(t) = 100 sin (250 t) V
i(t) = 20 cos (250 t) A
b) v(t) = 141.4 sin (500 t + 30°) V
i(t) = 0.707 cos (500 t - 60°) A
c) v(t) = 150 sin (2000 t - 45°) V
i(t) = 0.3 sin (2000 t + 45°) A
d) v(t) = 28 cos (100 t – 60°) V
i(t) = 3.535 sin (100 t – 60°) A
a) v(t) = 100 sin (250 t) V
i(t) = 20 cos (250 t) A = 20 sin (250 t +90°)
i leads v by 90°, the element is a capacitor.
XC =
P=
1.5
Vm 100
1
1
=
=5 Ω, C =
=
= 800 μF , 2
20
Im
ω X C ( 250)(5)
Vm I m
(100)(20)
cos(θ v − θ i ) =
cos(90°) = 0 W
2
2
1.5
b) v(t) = 141.4 sin (500 t + 30°) V
i(t) = 0.707 cos (500 t - 60°) A = 0.707 sin (500 t +30°)
i and v are in phase, the element is a resistor.
1.5
R=
Vm 141.4
=
= 200 Ω ,
I m 0.707
P=
Vm I m
(141.4)(0.707)
cos(θ v − θ i ) =
cos(0°) = 49.9849 W
2
2
2
1.5
c) v(t) = 150 sin (2000 t - 45°) V
i(t) = 0.3 sin (2000 t + 45°) A
i leads v by 90°, the element is a capacitor.
XC =
P=
1.5
Vm 150
1
1
=
= 500 Ω , C =
=
= 1 μF , 2
ω X C ( 2000)(500)
I m 0.3
Vm I m
(150)(0.3)
cos(θ v − θ i ) =
cos(90°) = 0 W
2
2
1.5
d) v(t) = 28 cos (100 t – 60°) V = 28 sin (100 t +30°) V
i(t) = 3.535 sin (100 t – 60°) A
v leads i by 90°, the element is an inductor.
XL =
P=
1.5
Vm
X
28
7.921
=
= 7.921 Ω , L = L =
= 79.21 mH , 2
100
ω
I m 3.535
Vm I m
( 28)(3.535)
cos(θ v − θ i ) =
cos(90°) = 0 W
2
2
Page (3 of 4)
1.5
EE308 Second Midterm Exam 1429/1430
Sunday 28/1/1430H
Question #4 (20 Marks)
For the network shown in Fig. 3 (frequency is 60 Hz),
(a) Determine I, VR, and VC in phasor form.
(b) Draw the phasor diagram of the voltages E, VR, and VC, and the current I.
(c) Calculate the total power factor, and indicate whether it is leading or lagging.
(d) Calculate the average power delivered to the circuit.
+
vR
+ vC
−
−
e = 2 (20) sin(377 t + 40°)
Fig. 3
a) X L1 = ω X L1 = (377)(0.2) = 75.4 Ω ,
X LT = 2 X L1 = (2)(75.4) = 150.8 Ω , 2
1
1
=
= 663.13 Ω 2
ω C (377)(4 × 10 −6 )
E = 20 ∠40° V
ZT = R + j XLT − j XC = 30 + j 150.8 − j 663.13 = 30 − j 512.33 Ω = 513.21 ∠86.65° Ω
E
20 ∠40°
=
= 38.97 ∠126.65° mA 3
I =
Z T 30 − j 512.33
XC =
VR = I R = (38.97 × 10 -3 ∠126.65°) (30) = 1.17 ∠126.65° V
3
VC = I (- j X C ) = (38.97 × 10 -3 ∠126.65°) (− j 663.13) = 25.84 ∠36.65° V
b)
3
3
VC
E
I
126.65°
40°
36.65°
VR
c) FP = cos(θV − θI ) = cos(40° – 126.65°) = cos( – 86.65°) = 0.0584, leading.
d) P = E I FP = (20) (38.97x 10-3) (0.0584) = 45.52 mW
Page (4 of 4)
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EE308 Second Midterm Exam 1429/1430
Sunday 28/1/1430H