1. Determine all of the angles between 0° and 360° whose sine is:
(a) 0.6792 (b) –0.1483
(a) Sine is positive in the 1st and 2nd quadrants If sin θ = 0.6792, then 1sin (0.6792)θ −= = 42.78° or 180° – 42.78° = 137.22° as shown in diagram (i) below.
(i) (ii) (b) Sine is negative in the 3rd and 4th quadrants If sin θ = 0.1483, then 1sin (0.1483)θ −= = 8.53° From diagram (ii) above, the two values where sin θ = –0.1483 are 180° + 8.53° = 188.53° and 360° – 8.53° = 351.47° 2. Solve the following equations for values of x between 0° and 360°:
(a) x = cos 1− 0.8739 (b) x = cos 1− (–0.5572)
(a) Cosine is positive in the 1st and 4th quadrants x = 1cos (0.8739)− = 29.08° or 360° –29.08° = 330.92° as shown in diagram (i) below.
(i) (ii) (b) Cosine is negative in the 2nd and 3rd quadrants x = 1cos (0.5572)− = 56.14° hence, from diagram (ii) shown above, the two values of x for which x = 1cos ( 0.5572)− − are: 180° – 56.14° = 123.86 and 180° + 56.14° = 236.14° 3. Find the angles between 0° to 360° whose tangent is:
(a) 0.9728 (b) –2.3420
(a) Tangent is positive in the 1st and 3rd quadrants 1tan 0.9728θ −= = 44.21° or 180° + 44.21° = 224.21° as shown in diagram (i) below.
(i) (ii) (b) Tangent is negative in the 2nd and 4th quadrants 1tan 2.3418θ −= = 66.88° and from diagram (ii) shown above, θ = 180° – 66.88° = 113.12° and 360° – 66.88° = 293.12° 4. Solve, in the range 0° to 360°, giving the answers in degrees and minutes: cos 1− (–0.5316) = t
Cosine is negative in the 2nd and 3rd quadrants.
1cos (0.5316)− = 57.886° or 57°53′ as shown in the diagram below.
From the diagram, t = 180° – 57°53′ = 122°7′ and t = 180° + 57°53′ = 237°53′ 5. Solve, in the range 0° to 360°, giving the answers in degrees and minutes: ( )1sin 0.6250 α− − =
Sine is negative in the 3rd and 4th quadrants If sin θ = 0.6250, then 1sin (0.6250)α −= = 38.682° = 38°41′
From the diagram above, the two values where sin α = –0.6250 are 180° + 38°41′ = 218°41′ and 360° – 38°41′ = 321°19′ 6. Solve, in the range 0° to 360°, giving the answers in degrees and minutes: tan 1− 0.8314 = θ
4. A sinusoidal voltage has a maximum value of 120 V and a frequency of 50 Hz. At time t = 0, the
voltage is (a) zero, and (b) 50 V. Express the instantaneous voltage v in the form v = A sin(ωt ± α).
Let v = A sin(ωt ± α) = 120 sin(2πft + φ) = 120 sin(100πt + φ) volts, since f = 50 Hz (a) When t = 0, v = 0 hence, 0 = 120 sin(0 + φ), i.e. 0 = 120 sin φ from which, sin φ = 0 and φ = 0 Hence, if v = 0 when t = 0, then v = 120 sin 100πt volts (b) When t = 0, v = 50 V hence, 50 = 120 sin(0 + φ)
from which, 50 sin120
φ= and 1 50sin 24.624 24.624 0.43rad120 180
πφ − = = °= × =
Hence, if v = 50 when t = 0, then v = 120 sin(100πt + 0.43) volts 5. An alternating current has a periodic time of 25 ms and a maximum value of 20 A. When
time = 0, current i = –10 amperes. Express the current i in the form i = A sin(ωt ± α).
If periodic time T = 25 ms, then frequency, 3
1 125 10
fT −
= =×
= 40 Hz
Angular velocity, ω = 2πt =2π(40) = 80π rad/s Hence, current i = 20 sin(80πt + φ) When t = 0, i = –10, hence –10 = 20 sin φ
from which, sin φ = 10 0.520
− = − and 1sin ( 0.5) 30 rad6
or πφ −= − = − ° −
Thus, 20sin 806
i t Aππ = −
or ( )20sin 80 0.524i t Aπ= −
6. An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At
time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin(ωt ± α).
Displacement, s = A sin(ωt ± α) where A = 3.2 m and 2 2 50 100fω π π π= = × =
i.e. s = 3.2 sin(100πt + α) At time t = 0, displacement = 150 cm = 1.5 m Hence, 1.5 = 3.2 sin(100πt + α) = 3.2 sin α
i.e. 1.53.2
= sin α
and α = 1 1.5sin3.2
−
= 27.953° = 0.488 rad
Hence, displacement = 3.2 sin(100πt + 0.488) m 7. The current in an a.c. circuit at any time t seconds is given by:
i = 5 sin(100πt – 0.432) amperes
Determine the (a) amplitude, frequency, periodic time and phase angle (in degrees)
(b) value of current at t = 0
(c) value of current at t = 8 ms
(d) time when the current is first a maximum
(e) time when the current first reaches 3 A.
Sketch one cycle of the waveform showing relevant points.
(a) If i = 5 sin(100πt – 0.432) mA, then amplitude = 5 A,
ω = 100π rad/s = 2πf from which, frequency, f = 1002ππ
= 50 Hz
periodic time, T = 1 150f
= = 0.020 s or 20 ms
and phase angle = 0.432 rad lagging or 1800.432π°
× = 24.75° lagging or 24°45′ lagging.
(b) When t = 0, i = 5 sin(–0.432) = –2.093 A (note that –0.432 is radians) (c) When t = 8 ms, i = 5 ( )3sin 100 8 10 0.432π − × − = 5 sin (2.081274) = 4.363 A (d) When the current is first a maximum, 5 = 5 sin(100πt – 0.432) i.e. 1 = sin(100πt – 0.432)
1. A complex current waveform i comprises a fundamental current of 50 A r.m.s. and frequency 100
Hz, together with a 24% third harmonic, both being in phase with each other at zero time. (a) Write
down an expression to represent current i. (b) Sketch the complex waveform of current using
harmonic synthesis over one cycle of the fundamental.
(a) Fundamental current: r.m.s. = 12
× maximum value
from which, maximum value = 2 r.m.s. 2 50× = × = 70.71 A Hence, fundamental current is: 1i = 70.71 sin 2π(100)t = 70.71 sin 628.3t A Third harmonic: amplitude = 24% of 70.71 = 16.97 A Hence, third harmonic current is: 3i = 16.97 sin 3(628.3)t = 16.97 sin 1885t A Thus, current i = 1i + 3i = (70.71 sin 628.3t + 16.97 sin 1885t) amperes (b) The complex waveform for current i is shown sketched below:
2. A complex voltage waveform v comprises a 212.1 V r.m.s. fundamental voltage at a frequency of
50 Hz, a 30% second harmonic component lagging the fundamental voltage at zero time by π/2 rad,
and a 10% fourth harmonic component leading the fundamental at zero time by π/3 rad. (a) Write
down an expression to represent voltage v. (b) Sketch the complex voltage waveform using
harmonic synthesis over one cycle of the fundamental waveform.
