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Typology: Assignments
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1 kHz supply. For each case calculate the resulting current flow and sketch the relevant phasor diagram.
R 2000 ; L 0.5 H; C 2.2 10 ^6 F; V 5V; f 103 Hz
Resistor V R
: volt so, mA
2 5. Ans
I^ V
V
I Capacitor: Similarly, since it is a perfect capacitor, then the only opposition to current flow is the capacitive reactance, XC.
X fC X V X
C
C
C
ohm
k
amp
so,
I 69. 1 A Ans
V
I
Inductor: Since this is a pure inductor, the only opposition to the flow of current will be the inductive reactance, XL. X fL X V X
L L
L
ohm k amp
so, m
I AA Ans
measured as 0.4 A. Determine the value of its inductance.
V 1 0V; f 200 Hz; I 0.4 A
X fL L X f
L L L L
ohm so, and, ohm so, henry
and, L 1. mH Ans
Calculate the capacitance value.
V 6V; f 5000 Hz; I 88.6 10 ^3 A
fC C
C C C
I ohm so, and, ohm
so,
fX C
C
farad
and F
.. Ans
from these two tests are as follows: d.c. supply of 1 0V; resulting current flow 50 mA a.c. supply of 1 0V, 1 00 Hz; resulting current flow 32 mA Using the results of these two tests, determine the resistance and inductance values for the coil.
d.c. test: V 1 0V; I 50 10 ^3 A
ฯ tan 1 X cos 1 sin^1 R
L L
In order to minimise possible errors the last of the above equations will be avoided, since it involves the use of two previously calculated values. So, the first equation has been chosen.
ฯ
ฯ
tan tan tan
and, or r
.. aad Ans
(c) (^) P V P
I cos watt cos so, W
ฯ 1 1
. Ans
Alternatively, since only resistive components dissipate power, then
P I^2 R^ watt 5 36. 2 25 7 8 2 1. W
Note : In this case the power cannot be calculated from P V I watt. This may be verifi ed by considering the circuit and phasor diagrams as shown below. From the circuit diagram it can be seen that the p.d. across the resistive component is VR and NOT V volt. This point illustrates the value of sketching the circuit and phasor diagrams before proceeding with the calculations.
I
I
R 25 ฮฉ
VR
V VR
V
V (^) L
VL 40 mH
ฯ
L
R R
volt V watt and, W, w
1 hhich vertifies the previous calculated answer.
the current flowing, (b) the p.d.s across the resistor and the capacitor, and (c) the circuit power factor.
R 270 ; C 10 ^5 F; V 20V; f 50 Hz The relevant circuit and phasor diagrams are shown below.
(a) X fC X Z R X
C
C C
5
2 2 2
ohm
so, ohm
2
and,
amp
hence, mA
I Ans
(b) V R V V X
R R C C
volt V volt
Ans (^)
VC. V Ans
(c) p.f. cos
so, p.f. lagging
ฯ R Z 270 4 7 4 0 347
. Ans
circuit is connected across a 1 00V, 80 Hz supply. Calculate (a) the circuit current, (b) the p.d.s. across the coil and the capacitor, (c) the circuit phase angle and power factor, and (d) the power dissipated.
R 330 ; L 0.25 H; C 10 ^5 F; V 1 00V; f 80 Hz
Note that we are dealing with a practical coil, which possesses both resistance and inductance. In order to simplify the calculations, such a coil is always considered as comprising a perfect resistor in series with a perfect inductor, as shown in the circuit diagram below.
I
I
R C 270 ฮฉ
V (^) R
VR
V (^) V 20 V
VC
VC
10 ฮผF (^) ฯ
(c) The complete phasor diagram is shown below.
I
V
Vcoil
VR
VL
(VC VL)
VC
ฯ
p.f. cos
hence, p.f. lagging phase
ฯ V V
Ans angle, cos so, lagging
ฯ ฯ
1 1
. Ans (d) P V P R
I cos ฯ watt or I^2 watt 00 0 296 0 977 2962 330 28
.9 9 W Ans P 28 9. W Ans
1 0V, variable frequency supply. Determine (a) the frequency at which the circuit current will be at its maximum value, (b) the value of this maximum current, and (c) the p.d.s across both the coil and the capacitor at this frequency.
For the current to be at its maximum value, the circuit must be supplied at its resonant frequency, fo Hz. This condition is shown by the phasor diagram below.
VL Vcoil
VR
VC
I
(a) (^) f LC f
o
o
Hz ^9 hence, kHz
. Ans
(b) At resonance, VL VC, so XL XC so they โcancelโ each other
and amp so, mA
20 Ans
(c) X^ X^ f L
C L o
C L C C
ohm
so, k ohm
hence, VC. V Ans
V Z Z R X
coil coil volt, where^ coil Lohm
and,
ccoil hence, (^) coil V
V. Ans