| Tutorials Home |
|
| | Alternating Current Theory |
|
| | More Engineering Links |
|
|
|
|
Examples on RMS, Peak Value, power dissipated
|
EXAMPLES : ---------
Example: 1 The equation of an alternating current is i=50 Ö2 x sin 400 pt ampere then the frequency and the root mean square of the current are respectively
(a) 200 Hz, 50 amp
(b) 400 p Hz,50 Ö2 amp
(c) 200 Hz, 50 Ö2 amp
(d) 50 Hz, 200 amp
Solution: (a) Comparing the given equation with i=i0 sin wt
Þ w = 400 p Þ 2 pn = 400 p Þ n = 200 Hz. Also irms = (iMsub>0 / Ö2) = (50 Ö2 /2) = 50A.
Example: 2 If the frequency of an alternating current is 50 Hz then the time taken for the change from zero to positive peak value and positive peak value to negative peak value of current are respectively
(a) 1/200 sec, 1/ 100 sec
(b) 1/ 100 sec, 1/200 sec
(c) 200 sec, 100 sec
(d) None of these
Solution: (a) Time take to reach from zero to peak value t= (T/4) = (1/4v) =(1/(4 x 50)) = (1/200) sec
Time take for the change from positive peak to negative peak t’ = (T/2) = (1/2v) = (1/(2 x 50 )) = (1/100) sec.
Example: 3 What will be the equation of ac of frequency 75 Hz if its r.m.s. value is 20 A
(a) i=20sin150 pt
(b) i=20 Ö2 sin (150 pt)
(c)i=(20/ Ö2) sin (150 pt)
(d)i=20 Ö2 sin (75 pt).
Solution: (b) By using i=i0 sin wt = i0 sin 2 pvt = irms Ö2 sin2pvt Þ I = 20 Ösin (150 pt)
Example: 4 At what time (From zero) the alternating voltage becomes(1/ Ö2) times of it's peak value. Where T is the periodic time
(a) (T/2)sec
(b) (T/4)sec
(c) (T/8) sec
(d) (T/12) sec
Solution: (c) By using V=V0 sin wt Þ (V0/ Ö2) = V0 sin (2 pt/T) Þ (1/ Ö2) = sin (2 p/T) t Þ sin (p/4) = sin (2 p/T) t
Þ( p/4) = (2 p/T)t Þ t = (T/8) sec
Example: 5 The peak value of an alternating e.m.f. E is given by E=E0cos w is 10 volts and its frequency is At timet=(1/600)sec the instantaneous e.m.f. is
(a) 10 V
(b) 5 ÖV
(c) 5 V
(d) 1V
Solution: (b) By usingE=E0 sin wt =10 cos 2 pn t = 10 cos 2 p ´ 50 ´(1/600) Þ E = 10 cos(p/6)=5 Ö3 V
Example: 6 The instantaneous value of current in an ac circuit is i=2 sin(100 pt + p/3) A The current at the beginning (t=0) will be
(a) 2 Ö3A
(b) Ö3 A
(c) (Ö3/2)A
(d) Zero
Solution: (b) At t = 0, i=2sin(0+( p/3)) = 2 x (Ö3/2) = Ö3 A.
Example: 7 The voltage of an ac source varies with time according to the equation V = 100 sin(100 pt) cos(100 pt) where t is in seconds and V is in volts. Then
(a) The peak voltage of the source is 100 volts
(b) The peak voltage of the source is 50 volts
(c) The peak voltage of the source is 100/ Ö2 volts
(d) The frequency of the source is 50 Hz
Solution: (b) The given equation can be written as follows
V=50 x 2sin 100 pt = 50sin2(100 pt)=50sin200 pt (\ sin 2 q = 2 sin q cos q )
Hence peak voltage V0 =50volt and frequency v= (200 p/2 p) = 100 Hz
Example: 8 If the frequency of ac is 60 Hz the time difference corresponding to a phase difference of 60o is
(a) 60 sec
(b) 1sec
(c) (1/60) sec
(d) (1/360) sec
Solution: (d) Time difference T.D.=(T/2 p) x Æ Þ T.D.=(T/2 p) x (p/3) =(T/6) = (1/6v) = (1/(6 x 60)) = (1/360) sec
Example: 9 In an ac circuit, V and i are given by V=100 sin (100 t) volt , and i= 100 sin (100 t + (p/3) ) mA . The power dissipated in circuit is
[MP PET 1989; MP PMT 1999]
(a) 104 watt
(b) 10watt
(c) 2.5 watt
(d) 5 watt
Solution: (c) P=(1/2)V0i0 cos Æ = (1/2) x (100 x 10-3 ) x cos (p/3) = 2.5 watt.
Example: 10 In a circuit an alternating current and a direct current are supplied together. The expression of the instantaneous current is given as i= 3+ 6 sin wt. Then the r.m.s. value of the current is
(a) 3A
(b) 6A
(c) 3 Ö2A
(d) 3 Ö3A
Solution: (d) The given current is a mixture of a dc component of 3A and an alternating current of maximum value 6A
Hence r.m.s. value = Ö{(dc)2 + (r.m.s. value of ac )2 ={(3)2 +(3 Ö2)2}=3 Ö3A.
Example: 11 The r.m.s. value of the alternating e.m.f. E = (8 sin w t + 6 sin 2 w t) V is
(a) 7.05 V
(b) 14.14 V
(c) 10 V
(d) 20 V
Solution: (a) Peak valueV0 = Ö{(8)2 + (6)2}=10 volt so vrms = (10/ Ö2)= 5 Ö2=7.05 volt
Example: 12 Voltage and current in an ac circuit are given by V=5 sin {100 pt – (p/6)} and i=4 sin {100 pt + (p/6)}
(a) Voltage leads the current by 30 °
(b) Current leads the voltage by 30 °
(c) Current leads the voltage by 60 °
(d) Voltage leads the current by 60 °
Solution: (c) Phase difference relative to current Δ Æ= -( p/6) – (p/6) =-( p/3)
In degree Δ Æ =-60° i.e. voltage lag behind the current by 60 ° or current leads the voltage by 60 °.
Example: 13 The instantaneous values of current and potential difference in an alternating circuit are i=sin wt and E = 100 cos w t respectively. r.m.s. value of wattless current (in amp) in the circuit is
(a) 1
(b) 1/ Ö2
(c) 100
(d) Zero
Solution: (b) r.m.s. value of wattless current=(i0/ Ö2) sin Æ
In this question i0 = 1 A and Æ (p/2). So r.m.s. value of wattless current =(1/ ÖA
Example: 14 The r.m.s. current in an ac circuit is 2 A. If the wattless current be Ö3A, what is the power factor
(a) 1/ Ö3
(b) 1Ö2
(c) 1/2
(d) 1/3
Solution: (c) iWL = irms sinÆ Þ Ö3 = 2 sinÆ =(Ö3/2) Þ Æ = 60° so p.f. = cos Æ cos 60° = 1/2
.
Example: 15 r.m.s. value of alternating current in a circuit is 4 A and power factor is 0.5. If the power dissipated in the circuit is 100W, then the peak value of voltage in the circuit is
(a) 50 volt
(b) 70 volt
(c) 35 volt
(d) 100 volt
Solution: (b) P=Vrms irms cosÆ Þ 100 = Vrms x 4 x 0.5 Þ Vrms = 50V so V0 = Ö2 x 50 =70 volt
Example: 16 The impedance of an ac circuit is 200 ohm and the phase angle between current and e.m.f is 60°. What is the resistance of the circuit
(a) 50 ohm
(b) 100 ohm
(c) 100Ö3 ohm
(d) 300 ohm
Solution: (b) By usingcos Æ (R/Z) Þ cos60° = (R/200) Þ 1/2 =R/200 Þ R= 100 ohm
Tricky example : 1
An ac voltage source of E = 150 sin 100 t is used to run a device which offers a resistance of 20 ohm and restricts the flow of current in one direction only. The r.m.s. value of current in the circuit will be
(a) 1.58 A
(b) 0.98 A
(c) 3.75 A
(d) 2.38 A
Solution : (c) As current flows in a single direction, the device allows current only during positive half cycle only
\ irms = (i0/2) = (V0 / 2R) = (150/(2 x 20)) = 3.75 A.
Tricky example :2
Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages
| Frequency in Hz | Phase lead of N over M in radians |
| (a) 0.4 | - p/4 |
| (b) 2.5 | - p/2 |
| (c) 2.5 | + p/2 |
| (d) 2.5 | - p/4 |
Solution : (b) From the graph shown below. It is clear that phase lead of N over M is –( p/2). Since time period (i.e. taken to complete one cycle) = 0.4 sec.
Hence frequency v=(1/T) = 2.5 Hz
|
|
|
