Module
4
Single-phase AC circuits
Version 2 EE IIT, Kharagpur
Lesson
16
Solution of Current in
AC Parallel and Seriesparallel Circuits
Version 2 EE IIT, Kharagpur
In the last lesson, the following points were described:
1. How to compute the total impedance/admittance in series/parallel circuits?
2. How to solve for the current(s) in series/parallel circuits, fed from single phase ac
supply, and then draw complete phasor diagram?
3. How to find the power consumed in the circuit and also the different components, and
the power factor (lag/lead)?
In this lesson, the computation of impedance/admittance in parallel and series-parallel
circuits, fed from single phase ac supply, is presented. Then, the currents, both in
magnitude and phase, are calculated. The process of drawing complete phasor diagram is
described. The computation of total power and also power consumed in the different
components, along with power factor, is explained. Some examples, of both parallel and
series-parallel circuits, are presented in detail.
Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power
factor.
After going through this lesson, the students will be able to answer the following
questions;
1. How to compute the impedance/admittance, of the parallel and series-parallel circuits,
fed from single phase ac supply?
2. How to compute the different currents and also voltage drops in the components, both
in magnitude and phase, of the circuit?
3. How to draw the complete phasor diagram, showing the currents and voltage drops?
4. How to compute the total power and also power consumed in the different
components, along with power factor?
This lesson starts with two examples of parallel circuits fed from single phase ac
supply. The first example is presented in detail. The students are advised to study the two
cases of parallel circuits given in the previous lesson.
Example 16.1
The circuit, having two impedances of Z1 = (8 + j 15) Ω and Z 2 = (6 − j 8) Ω in
parallel, is connected to a single phase ac supply (Fig. 16.1a), and the current drawn is 10
A. Find each branch current, both in magnitude and phase, and also the supply voltage.
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I1
Z1 = (8 + j15) Ω
A
B
I = 10A
I2
Z2 = (6 – j8) Ω
Fig. 16.1 (a) Circuit diagram
Solution
Z1 ∠φ1 = (8 + j 15) = 17 ∠61.93° Ω
Z 2 ∠ − φ 2 = (6 − j 8) = 10 ∠ − 53.13° Ω
I ∠0° (OC ) = 10 ∠0° = (10 + j 0) A
The admittances, using impedances in rectangular form, are,
8 − j 15 8 − j 15
1
1
=
= 2
=
= ( 27.68 − j 51.9) ⋅ 10 −3 Ω −1
Y1 ∠ − φ1 =
289
Z 1 ∠φ1 8 + j 15 8 + 15 2
6 + j8 6 + j8
1
1
Y2 ∠φ 2 =
=
= 2
=
= (60.0 + j 80.0) ⋅ 10 −3 Ω −1
2
Z 2 ∠ − φ2 6 − j 8 6 + 8
100
Alternatively, using impedances in polar form, the admittances are,
1
1
Y1 ∠ − φ1 =
=
= 0.05882 ∠ − 61.93°
Z 1 ∠φ1 17.0 ∠61.93°
= (27.68 − j 51.9) ⋅ 10 −3 Ω −1
1
1
Y2 ∠φ 2 =
=
= 0.1 ∠53.13° = (60.0 + j 80.0) ⋅ 10 −3 Ω −1
Z 2 ∠ − φ 2 10.0 ∠ − 53.13°
The total admittance is,
Y ∠φ = Y1 + Y2 = [(27.68 − j 51.9) + (60.0 + j 80.0)] ⋅ 10 −3 = (87.68 + j 28.1) ⋅ 10 −3
= 92.07 ⋅ 10 −3 ∠17.77° Ω −1
The total impedance is,
1
1
Z ∠ −φ =
=
= 10.86 ∠ − 17.77° = (10.343 − j 3.315) Ω
Y ∠φ 92.07 ⋅ 10 −3 ∠17.77°
The supply voltage is
V ∠ − φ (V AB ) = I ∠0° ⋅ Z ∠ − φ = (10 × 10.86) ∠ − 17.77° = 108.6 ∠ − 17.77° V
= (103.43 − j 33.15) V
The branch currents are,
V ∠ − φ ⎛ 108.6 ⎞
I 1 ∠ − θ 1 (OD) =
=⎜
⎟ ∠ − (17.77° + 61.93°) = 6.39 ∠ − 79.7° A
Z1 ∠φ1 ⎝ 17.0 ⎠
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= (1.143 − j 6.286 ) A
I 2 ∠θ 2 (OE ) = I ∠0° − I1 ∠ − θ1 (OC − OD = OC − CE )
= (10.0 + j 0.0) − (1.143 − j 6.286) = (8.857 + j 6.286) A = 10.86 ∠35.36° A
Alternatively, the current I 2 is,
I 2 ∠θ 2 (OE ) =
V ∠ − φ ⎛ 108.6 ⎞
=⎜
⎟ ∠(−17.77° + 53.13°) = 10.86 ∠35.36° A
Z 2 ∠ − φ2 ⎝ 10.0 ⎠
= (8.857 + j 6.285) A
The phasor diagram with the total (input) current as reference is shown in Fig. 16.1b.
E
I2= 10.86
53.13°
θ1 = 35.35
O
θ1 = 79.7°
Ф = 17.8°
C
I = 10A
61.90
VAB
I1 = 6.4 A
108.63 V
D
Fig. 16.1 (b) Phasor diagram
Alternative Method
Z ′ ∠φ ′ = Z1 ∠φ1 + Z 2 ∠ − φ 2 = (8 + j 15) + (6 − j 8) = (14 + j 7) = 15.65 ∠26.565° Ω
Z ⋅Z
Z ∠φ ⋅ Z ∠ − φ 2 ⎛ 17.0 × 10.0 ⎞
Z ∠ −φ = 1 2 = 1 1 2
=⎜
⎟ ∠(61.93° − 53.13° − 26.565°)
Z1 + Z 2
Z ′ ∠φ ′
⎝ 15.65 ⎠
= 10.86 ∠ − 17.77° = (10.343 − j 3.315) Ω
The supply voltage is
V ∠ − φ (VAB ) = I ⋅ Z = (10 ×10.86) ∠ − 17.77° = 108.6 ∠ − 17.77° V
= (103.43 − j 33.15) V
The branch currents are,
Z2
⎛ 10.0 × 10.0 ⎞
I1 ∠ − θ1 (OD) = I
=⎜
⎟ ∠(−53.13° − 26.565°) = 6.39 ∠ − 79.7° A
Z1 + Z 2 ⎝ 15.65 ⎠
= (1.142 − j 6.286) A
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I 2 ∠θ 2 (OE ) = I − I1 (OC − OD = OC − CE ) = (10.0 + j 0.0) − (1.142 − j 6.286)
= (8.858 + j 6.286) A = 10.862 ∠35.36° A
Alternatively, the current I 2 is,
Z1
⎛ 10.0 × 17.0 ⎞
=⎜
I 2 ∠θ 2 (OE ) = I
⎟ ∠(61.93° − 26.565°) = 10.86 ∠35..36° A
Z1 + Z 2 ⎝ 15.65 ⎠
= (8.858 + j 6.286) A
Example 16.2
The power consumed in the inductive load (Fig. 16.2a) is 2.5 kW at 0.71 lagging
power factor (pf). The input voltage is 230 V, 50 Hz. Find the value of the capacitor C,
such that the resultant power factor of the input current is 0.866 lagging.
IL
I
+
IC
C
-
L
O
A
D
230 V
Fig. 16.2 (a) Circuit diagram
Solution
P = 2.5 KW = 2.5 ⋅ 10 3 = 2500 W
V = 230 V
f = 50 Hz
The power factor in the inductive branch is cos φ L = 0.71 (lag )
The phase angle is φ L = cos −1 (0.71) = 44.77° ≈ 45°
P = V ⋅ I L cos φ L = 230 ⋅ (I L cos φ L ) = 2500
P
2500
IL =
=
= 15.31 A
V cos φ L 230 × 0.71
I L cosφ L = 15.31× 0.71 = 10.87 A ; I L sin φ L = 15.32 × sin 45° = 10.87 A
The current I L is, I L ∠ − φ L = 15.31∠ − 45° = (10.87 − j 10.87) A
The power consumed in the circuit remains same, as the capacitor does not consume
any power, but the reactive power in the circuit changes. The active component of the
total current remains same as computed earlier.
I cos φ = I L cos φ L = 10.87 A
The power factor of the current is cos φ = 0.866 (lag )
The phase angle is φ = cos −1 (0.866) = 30°
The magnitude of the current is I = 10.87 / 0.866 = 12.55 A
The current is I ∠ − φ = 12.55 ∠ − 30° = (10.87 − j 6.276) A
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The current in the capacitor is
I C ∠90° = I ∠ − φ − I L ∠ − φ L = (10.87 − j 6.276) − (10.87 − j 10.87)
= j 4.504 = 4.504 ∠90° A
This current is the difference of two reactive currents,
− I C = I sin φ − I L sin φ L = 6.276 − 10.87 = −4.504 A
1
230
V
The reactance of the capacitor, C is X C =
=
=
= 51.066 Ω
2 π f C I C 4.504
1
1
The capacitor, C is C =
=
= 62.33 ⋅ 10 −6 = 62.33 μF
2 π f X C 2 π × 50 × 51.066
The phasor diagram with the input voltage as reference is shown in Fig. 16.2b.
IC
4.5 A
230 V
A
45°
φ
C
IC
15.3
B
Fig. 16.2 (b) Phasor diagram
Example 16.3
An inductive load (R in series with L) is connected in parallel with a capacitance C of
12.5 μF (Fig. 16.3a). The input voltage to the circuit is 100 V at 31.8 Hz. The phase
angle between the two branch currents, ( I 1 = I L ) and ( I 2 = I C ) is 120° , and the current
in the first branch is I 1 = I L = 0.5 A . Find the total current, and also the values of R & L.
I
A
I2
100 V
+
-
R
D
L
I 1 = 0.5A
C = 12.5 μF
B
Fig. 16.3 (a) Circuit diagram
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Solution
V = 100 V
ω = 2 π f = 2 π × 31.8 ≈ 200 rad / s
−6
I 1 = 0.5 A
C = 12.5 μF = 12.5 ⋅ 10 F
X C = 1 /(ω C ) = 1 /( 200 × 12.5 ⋅ 10 −6 ) = 400 Ω
The current in the branch no. 2 is
I 2 ∠90° = V /(− j X C ) = 100 ∠0° / 400 ∠ − 90° = (100 / 400) ∠90° = 0.25 ∠90°
= (0.0 + j 0.25) A
The current in the branch no. 1 is I 1 ∠ − φ1 = 0.5 ∠ − φ1
The phase angle between I 1 and I 2 is 90° + φ1 = 120°
So, φ1 = 120° − 90° = 30°
I 1 ∠ − 30° = 0.5 ∠ − 30° = (0.433 − j 0.25) A
The impedance of the branch no. 1 is,
Z 1 ∠φ1 = ( R + j X L ) = V ∠0° / I 1 ∠ − 30° = (100 / 0.5) ∠30° = 200 ∠30°
= (173.2 + j 100.0) Ω
X L = ω L = 100.0 Ω
R = 173.2 Ω
So, L = X L / ω = 100 / 200 = 0.5 H = 500 ⋅ 10 −3 = 500 mH
The total current is,
I ∠0° = I 1 ∠ − 30° + I 2 ∠90° = (0.433 − j 0.25) + j 0.25 = (0.433 + j 0.0)
= 0.433 ∠0° A
The total impedance is,
Z ∠0° = ( R ′ + j 0) = V ∠0° / I ∠0° = (100 / 0.433) ∠0° = 231.0 ∠0°
= (231.0 + j 0.0) Ω
The current, I is in phase with the input voltage, V .
The total admittance is Y ∠0° = Y1 ∠ − φ1 + Y2 ∠90° = (1 / Z 1 ∠30°) + (1 / Z 2 ∠ − 90°)
The total impedance is Z ∠0° = ( Z1 ∠30° ⋅ Z 2 ∠ − 90°) /( Z1 ∠30° + Z 2 ∠ − 90°)
Any of the above values can be easily calculated, and then checked with those
obtained earlier. The phasor diagram is drawn in Fig. 16.3b.
f = 31.8 Hz
Solution of Current in Series-parallel Circuit
Series-parallel circuit
The circuit, with a branch having impedance Z 1 , in series with two parallel branches
having impedances, Z 2 and Z 3 , shown in Fig. , , is connected to a single phase ac
supply.
The impedance of the branch, AB is Z AB ∠φ AB = Z1 ∠φ1
1
1
; Y3 ∠ − φ 3 =
Y2 ∠ − φ 2 =
Z 2 ∠φ 2
Z 3 ∠φ 3
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The admittance of the parallel branch, BC is
1
1
YBC ∠ − φ BC = Y2 ∠ − φ 2 + Y3 ∠ − φ3 =
+
Z 2 ∠φ 2 Z 3 ∠ φ 3
The impedance of the parallel branch, BC is
Z ∠φ 2 + Z 3 ∠φ 3 Z 2 ∠φ 2 + Z 3 ∠φ 3
1
Z BC ∠φ BC =
= 2
=
(Z 2 Z 3 ) ∠(φ 2 + φ 3 )
YBC ∠ − φ BC
Z 2 ∠φ 2 ⋅ Z 3 ∠φ 3
The total impedance of the circuit is
Z AC ∠φ AC = Z AB ∠φ AB + Z BC ∠φ BC = Z 1 ∠φ1 + Z BC ∠φ BC
The supply current is
V ∠ 0°
I ∠ − φ AC =
Z AC ∠φ AC
The current in the impedance Z 2 is
Z 3 ∠φ 3
I 2 ∠φ 4 = I ∠ − φ AC
(Z 2 ∠φ 2 + Z 3 ∠φ3 )
Thus, the currents, along with the voltage drops, in all branches are calculated. The
phasor diagram cannot be drawn for this case now. This is best illustrated with the
following examples, where the complete phasor diagram will also be drawn in each case.
I2 = 0.25
I = 0.433
100 V B
30°
86.6 V
50 V 0.25 I2
D
Z1
0.5 I1
A
Fig. 16.3 (b) Phasor diagram
Example 16.4
Find the input voltage at 50 Hz to be applied to the circuit shown in Fig. 16.4a, such
that the current in the capacitor is 8 A?
I1
R1 = 5 Ω
I
A
•
•
L1 = 25.5 mH
I1
I 2 = 8A
R2 = 8 Ω
•D
R3 = 7 Ω
L3 = 38.2 mH
•B
318 μF
Fig. 16.4 (a) Circuit diagram
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Solution
ω = 2 π f = 2 π × 50 = 314.16 rad / s
L1 = 0.0255 H
L3 = 0.0382 H
C = 318 μF = 318 ⋅ 10 −6 F
X 1 = ω L1 = 314.16 × 0.0255 = 8 Ω
X 3 = ω L3 = 314.16 × 0.0382 = 12 Ω
f = 50 Hz
I 2 ∠0° = 8 ∠0° = (8 + j 0) A
X 2 = 1 /(ω C ) = 1 /(314.16 × 318 ⋅ 10 −6 ) = 10 Ω
Z1 ∠φ1 = R1 + j X 1 = (5 + j 8) = 9.434 ∠58° Ω
Z 2 ∠ − φ 2 = R2 − j X 2 = (8 − j 10) = 12.806 ∠ − 51.34° Ω
Z 3 ∠φ 3 = R3 + j X 3 = (7 + j 12) = 13.89 ∠59°.74 Ω
V AC ∠ − φ 2 = I 2 ∠0° ⋅ Z 2 ∠ − φ 2 = (8.0 × 12.806) ∠ − 51.34° = 102.45 ∠ − 51.34° V
= (64 − j 80) V
V AC ∠ − φ 2 ⎛ 102.45 ⎞
=⎜
⎟ ∠ − (51.34° + 58°) = 10.86 ∠ − 109.34° A
Z 1 ∠φ1
⎝ 9.434 ⎠
= (−3.6 − j 10.25) A
I ∠ − θ 3 = I 1 ∠ − θ 1 + I 2 ∠0° = −(3.6 + j 10.25) + (8.0 + j 0.0) = (4.4 − j 10.25)
I1 ∠ − θ1 =
= 11.154 ∠ − 66.77° A
VCB ∠ − θ CB = I ∠ − θ 3 ⋅ Z 3 ∠φ 3 = (11.154 × 13.89) ∠(−66.77° + 59.74°)
= 154.93 ∠ − 7.03° V = (153.764 − j 18.96) V
V AB ∠ − θ AB = V AC ∠ − φ 2 + VCB ∠ − θ CB = (64.0 − j 80.0) + (153.764 − j 18.96)
= (217.764 − j 9 0.96) = 239.2 ∠ − 24.44° V
The phasor diagram with the branch current, I 2 as reference, is shown in Fig. 16.4b.
I2 = 8A
7.03°
51.34°
VDB = 154.9V
66.77°
58°
VAD = 102.45 V
I
I1 = 10.86A
VAB = 239.2
11.15A
Fig. 16.4 (b) Phasor diagram
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Example 16.5
A resistor of 50 Ω in parallel with an inductor of 30 mH, is connected in series with a
capacitor, C (Fig. 16.5a). A voltage of 220 V, 50 Hz is applied to the circuit. Find,
(a) the value of C to give unity power factor,
(b) the total current, and
(c) the current in the inductor
R = 50 Ω
A
•
•
D
B
C
L = 30 mH
230 V
50 Hz
Fig. 16.5 (a) Circuit diagram
Solution
ω = 2 π f = 2 π × 50 = 314.16 rad / s
V = 220 V
L = 30 mH = 30 ⋅ 10 −3 = 0.03 H
R = 50 Ω
X L = ω L = 314.16 × 0.03 = 94.24 Ω
f = 50 Hz
The admittance, Y AD is,
1
1
1
1
Y AD ∠ − φ AD = +
=
+
= (20.0 − j 10.61) ⋅ 10 −3
R j X l 50 j 94.24
= 0.02264 ∠ − 27.95° Ω −1
The impedance, Z AD is,
Z AD ∠φ AD = 1 /(Y AD ∠φ AD ) = 1 /(0.02264 ∠ − 27.95°) = 44.17 ∠27.95°
= (39.02 + j 20.7) Ω
The impedance of the branch (DB) is Z DB = − j X C = − j [1 /(ω C ) .
As the total current is at unity power factor (upf), the total impedance, Z AB is
resistive only.
Z AB ∠0° = R AB + j 0 = Z AD ∠φ AD + Z DB ∠ − 90° = 39.02 + j ( 20.7 − X C )
Equating the imaginary part, X c = 1 /(ω C ) = 20.7 Ω
The value of the capacitance C is,
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1
1
=
= 153.8 ⋅ 10 −6 = 153.8 μF
ω X C 314.16 × 20.7
So, Z AB ∠0° = R AB + j 0 = (39.02 + j 0.0) = 39.02 ∠0° Ω
The total current is,
I ∠0° = V ∠0° / Z AB ∠0° = (220.0 / 39.02) ∠0° = (5.64 + j 0.0) = 5.64 ∠0° A
The voltage, V AD is,
VAD ∠φ AD = I ∠0° ⋅ Z AD ∠φ AD = (5.64 × 44.17) ∠27.95° = 249.05 ∠27.95° V
C=
= (220.0 + j 116.73) V
The current in the inductor, I L is,
I L ∠θ L = V AD ∠φ AD / X L ∠90° = (249.05 / 94.24) ∠(27.95° − 90°)
= 2.64 ∠ − 62.05° A = (1.24 − j 2.335) A
The phasor diagram is shown in Fig. 16.5b.
249.0
D
116.7 V
28°
IR = 4.98
28°
62°
A
5.64
B
220 V
116.7 V
IL = 2.64 A
(i)
(ii)
Fig. 16.5 (b) Phasor diagram
Example 16.6
In the circuit (Fig. 16.6a) the wattmeter reads 960 W and the ammeter reads 6 A.
Calculate the values of VS , VC , I C , I , I L and X C .
I
A
•
+
W
A
R1 = 10 Ω
R2 = 6 Ω
D
•
E
•
IL
VC
IC C
IC
jxL = j 8 Ω
-
•B
Fig. 16.6 (a) Circuit diagram
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Solution
In this circuit, the power is consumed in two resistance, R1 and R2 only, but not
consumed in inductance L, and capacitance C. These two components affect only the
reactive power.
P = 960 W
R2 = 6 Ω
I=6A
XL =8 Ω
R1 = 10 Ω
Total power is, P = I 2 ⋅ R1 + I L2 ⋅ R2 = (6) 2 × 10 + 6 ⋅ I L2 = 360 + 6 ⋅ I L2 = 960 W
or, 6 ⋅ I L2 = 960 − 360 = 600 W
So, I L = 600 / 6 = 10 A
The impedance of the inductive branch is,
Z L = R2 + j X L = (6 + j 8) = 10 ∠53.13° Ω
The magnitude of the voltage in the inductive branch is,
V DB = VC = I L ⋅ Z L = 10 × 10 = 100 V
Assuming VDB = 100 ∠0° as reference, the current, I L is,
I L ∠ − φ L = VDB ∠0° / Z L ∠φ L = (100 / 10) ∠ − 53.13° = 10 ∠ − 53.13° A = (6 − j 8) A
The current, I C ∠90° = j I C = VDB ∠0° / X C ∠ − 90° = (V / X C ) ∠90°
The total current is I ∠φ = I L ∠ − φ L + I C ∠90° = (6 − j 8) + j I C = 6 + j ( I C − 8)
So, I = (6) 2 + ( I C − 8) 2 = 6 A
or, 36 + ( I C − 8) 2 = (6) 2 = 36
So, I C = 8 A
The capacitive reactance is, X C = VDB / I C = 100 / 8 = 12.5 Ω
The total current is I ∠0° = (6 + j 0) = 6 ∠0° A
or, it can be written as,
I ∠ 0 ° = I L ∠ − φ L + I C ∠90 ° = ( 6 − j 8) + j 8 = ( 6 + j 0 ) = 6 ∠ 0 ° A
The voltage V AD is.
V AD ∠0° = I ∠0° ⋅ ( R1 + j 0) = ( I ⋅ R1 ) ∠0° = (6 × 10) ∠0° = 60 ∠0° = (60 + j 0) V
The voltage, VS = V AB is.
VS = V AB ∠0° = V AD ∠0° + VDB ∠0° = (60 + 100) ∠0° = 16 0 ∠0° = (160 + j 0) V
The current, I is in phase with VS = V AB , and also VDB .
The total impedance is,
Z AB ∠0° = Z AD ∠0° + Z DB ∠0° = V AB ∠0° / I ∠0° = (160 / 6) ∠0° = 26.67 ∠0°
= (26.67 + j 0.0) Ω
The impedance, Z DB is,
Z DB ∠0° = Z AB ∠0° − Z AD ∠0° = VDB ∠0° / I ∠0° = (100 / 6) ∠0° = 16.67 ∠0°
= (16.67 + j 0.0) Ω
Both the above impedances can be easily obtained using the circuit parameters by the
method given earlier, and then checked with the above values. The impedance, Z DB can
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be obtained by the steps given in Example 16.3. The phasor diagram is shown in Fig.
16.6b.
8A
A
60 V
I = 6A D
53.13° A
B
6A
AB = 160 V (VO)
8A
Fig. 16.6 (b) Phasor diagram
Starting with the examples of parallel circuits, the solution of the current in the seriesparallel circuit, along with the examples, was taken up in this lesson. The problem of
resonance in series and parallel circuits will be discussed in the next lesson. This will
complete the module of single phase ac circuits
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Problems
16.1 Find the impedance, Zab in the following circuits (Fig. 16.7a-b): (check with
admittance diagrams in complex plane)
a •
•
a •
•
l
l
l
-j2
b •
j1
- j1
jl
•
b•
•
(b)
(a)
Fig. 16.7
16.2
A resistor (R) of 50 Ω in parallel with a capacitor (C) of 40 μF, is connected in
series with a pure inductor (L) of 30 mH to a 100 V, 50 Hz supply. Calculate the
total current and also the current in the capacitor. Draw the phasor diagram.
16.3
In a series-parallel circuit (Fig.16.8), the two parallel branches A and B, are in
series with the branch C. The impedances in Ω are, ZA = 5+j, ZB = 6-j8, and ZC
= 10+j8. The voltage across the branch, C is (150+j0) V. Find the branch
currents, IA and IB, and the phase angle between them. Find also the input
voltage. Draw the phasor diagram.
16.4
A total current of 1A is drawn by the circuit (Fig.16.9) fed from an ac voltage, V
of 50 Hz. Find the input voltage. Draw the phasor diagram.
ZA = (5 + j6) Ω
ZC = (10 + j8) Ω
•
Q
P •
ZB = (6 – j8) Ω
•R
150 V
Fig. 16.8
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R1 = 25 Ω
B
A
I=1A
+
60 μF
V
-
R2 = 30 Ω
L = 50 mH
Fig. 16.9
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List of Figures
Fig. 16.1 (a) Circuit diagram (Ex. 16.1)
(b) Phasor diagram
Fig. 16.2 (a) Circuit diagram (Ex. 16.2)
(b) Phasor diagram
Fig. 16.3 (a) Circuit diagram (Ex. 16.3)
(b) Phasor diagram
Fig. 16.4 (a) Circuit diagram (Ex. 16.4)
(b) Phasor diagram
Fig. 16.5 (a) Circuit diagram (Ex. 16.5)
(b) Phasor diagram
Fig. 16.6 (a) Circuit diagram (Ex. 16.6)
(b) Phasor diagram
Version 2 EE IIT, Kharagpur