QUESTION
1. FIGURE 1 shows the positions of four towns marked on a map.
W
32km
3
θ
58°
?
?
27km
Z
FIG. 1
X
(a) Calculate, to the nearest kilometre, the distance between towns W
and Y.
(b) Calculate, to the nearest degree, the size of the angle marked
θ.
Y
WX = 32km
WZ = 27km
XYW = 58°
(c) Calculate, to the nearest kilometre, the distance between towns Y and
Z.
2. The instantaneous value of current, i amp, at t seconds is given by:
Find the value of
(a) The amplitude.
(b) The period.
(c) The frequency.
(d) The initial phase angle (when t = 0), expressed in both radians and
degrees.
(e) The value of i when t = 2.5 s.
(f) The time (in milliseconds) when the current first reaches its maximum
value.
3. Use the trigonometrical identities to expand and simplify if possible.
(a) (i) cos(270° –
θ)
(ii) sin(270° –
θ)
(iii) cos(270° +
θ)
(b) The voltage V
If V
3
1
and V
is the sum of V
Find the expression of V
1
2
are represented by as follows,
and V
3
2
,
in sine waveform
and verify that the resultant voltage V
and V
2
.
3
is in the same frequency as V
4. (a) Given the equivalent impedance of a circuit can be calculated by the
expression
If Z
1
= 4 + j10 and Z
= 12 – j3, calculate the impedance Z in
both rectangular and polar forms.
2
(b) Three impedances are connected in parallel, Z
= 2 + j2,
Z
2
= 1 + j5 and Z
where:
3
Vt
1
Vt
2
VVV
312
VR t
3
5
=
()
3
=
()
2
sin
cos
=+
=+
()
sin ωα
Z
ZZ
12
12
ZZ
=
+
ω
ω
= j6. Find the equivalent admittance Y.
=++
111
Y
ZZZ
123
Express the admittance in both rectangular and polar forms.
5. The potential difference across a circuit is given by the complex number
Vj=+40 35 volts
and the current is given by the complex number
Ij=+6 3 amps
(a) Sketch the appropriate phasors on an Argand diagram.
(b) Find the phase difference (i.e. the angle
φ) between the phasors for
voltage V and the current I.
(c) Find the power (watt), given that
Teesside University Open Learning
(Engineering)
6
power = VI××
SOLUTION
1) a) Let Nearest distance between W and Y = d
From Triangle WXY, Sin(58 ) = WX/ WY
Sin(58) = 32/d
d = 32/sin(58)
= 32/0.848 = 37.74 Km Answer.
b) From Triangle WYZ ,
Cos (theta) = WZ/WY
= 27/37.74
= 0.715
Theta = arcos(0.715)
Theta = 44.32 degree Answer
c) From Triangle WYZ , tan(theta) = YZ/WZ
tan(44.32) = YZ /27
YZ = 27* tan(44.32) = 27*0.9766 = 26. 37 Km Answer
2. a) Amplitude = Maximum displacement rom mean position =- 15 Answer
b) Period = ?
w (omega) = 2 pi f = 100 pi
f = 100 pi/2 pi = 50
Now,Period T = 1/f = 1/50 = 0.02 s
= 20 milliseconds Answer.
C) Frequency f = 50 Hz Answer
d) Initial phase angle (when t = 0 ) = 0.6 radians
Initial phase angle in degree = 0.6 *180/3.14
= 34.377 degrees Answer
e) When t = 2.5 s
i = 15 sin( 100*3.14*2.5 +0.6)
= 15 sin ( 250 *pi +0.6)
= 15 sin(0.6) = 15*0.5646 = 8.47 Amp Answer
f) maximum value of sine function = 1
So, Current I is maximum when sine function is maximum
Hence sin (100 pi t + 0.6) = 1
100 pi t +0.6 = pi/2 = 90 degree
314 t + 34.377 = 90
314t = 90 – 34.377
t = 55.62/314 = 0.177
time (in milliseconds) t = 177.14 milli seconds Answer
3. a) i) cos(270 – theta) = cos( 180 +90 – theta) = – cos(90-theta) = – sin (theta) Answer
ii) sin(270 – theta) = sin(180 +90-theta) = -sin(90 –theta) = – cos(theta) Answer
iii) cos(270 +theta) = cos(180 +90 +theta) = – cos( 90 + theta )
= – ( -sin (theta)) = sin (theta) Answer
b) V3 = v1 +v2
= 3 sin(wt) + 2cos(wt)
= sqrt (13){ 3/sqrt13 sin(wt) + 2/sqrt13 cos (wt)}
= sqrt(13) { sin (wt + alpha) }
Here R = sqrt(13) ,alpha = arccos ( 3/sqrt13)
Here frequency of V1 = w /2pi
Frequency of v2 = w/2pi Frequency of V3 = w/2 pi Answer
4. a)
Z = z1.z2 / (Z1+Z2)
Z1.z2 = (4+j10)(12-j3) = 48 +30 = 78 + j108
Z1+z2 = 4+j10 +12 –j3 = 16 +j7
Z = (78 +j 108)/(16 +j7)
Z = (78+j108)( 16 – j7)/ (305)
Z = { (1248 +756) +j (1728+546) }/305
= (2004+ j2274)/ 305
Rectangular form Z = 6.57 + j 7.46 Answer
Polar form z = 9.94 {cos(46.63) +j sin(46.63)}
b) 1/z1 = 1/(2+j2) = 0.25 – j0.25
1/z2 = 1/(1+j5) = 0.038 – j 0.19
1/z3 = 1/j6 = – j0.167
Y = 0.25 –j0.25 +0.038 –j0.19 – j0.167
= 0.288 – j 0.646 Answer
Rectangular form
Polar form = 0.707{ cos( 65.96) – j sin(65.96) }
5 . a) Vphase (alpha ) = 41.186 and Iphase (beta) = 26.565. To draw voltage given V function. we take 40 uint on x plane and 35 unit on y plane.and connect the intersection point to origin .The angle in positive x axis direction gives phase.
Similarly to draw I function we take 6 unit on x –plane and 3 unit on y plane and ollow same method to find phase of I.
b) Phase dierence (Phi) = 41.186 -26.565 = 14.621
c ) mod of v = 53.15 , mod of I = 6.707 and cos (14.621) = 0.9676
Power P = 53.15*6.707 *0.9676 = 344.93 watts
GF35
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