Now a-days, owing to multiple system of transmission and distribution, we come across par allel circuits (i.e., impedances joined in parallel) more often. Practically all lighting and power circuits are constant voltage circuits with the’ loads connected in parallel. In a parallel A. C. circuit (like parallel D.C. circuit) the voltage is the same across each brand L
4.6.2. Methods for solving A. C. parallel circuits
The following three methods are available to solve such circuits:
1. Phasor or vector method
2. Admittance method
3. Vector algebra (symbolic method or )·method)
1. Vector or phasor method:
Consider a parallel Circuit consisting of two b:-anches of impedances Z1(R1, L) and Z2(R2, C) respectively, and connected in parallel across an alternating voltage V volts (r.m.s.), as shown in Fig. 54 (a). Since the two branches are connected in parallel therefore, the voltage across each branch is the same and equal to supply voltage V but currents through them will be different.
Fig. 54. Single -Phase parallel circuit.- Phasor method.
Braneh-1 Impedance, Z 1 = R1 + XL
Current, I 1 = V / Z1
Power factor , cos ϕ 1 = R1 / Z1 or ϕ 1 = cos -1 (R1 / Z1)
Current, J1 1ags behind the applied voltage by $1
Braneh-2 Impedance, Z2 = R2 + ZC
Current, I2 = V / Z2
Power factor , cos ϕ2 = R2 / Z2 or ϕ2 = cos -1 (R2 / Z2)
Currenti2 1eads V by ϕ2 (Fig. 54- (b)]
Resultant current I , which is phasor sum of I 1 and I 2, Can be determined either by using parallelogram law or phasors, as shown in Fig. 54 (b) or by resolving branch currents I 1 and I 2 along X-axis and Y-axis and then determining the resultant of these components analytically.
Component of resultant current I along X-axis
= Sum of components of branch currents I 1 and I 2 along X-axis
I cos ϕ = I1 cos ϕ2 + 12 cos ϕ2
Similarly, component of resultant current I along Y-axis
= Sum of components of branch currents / 1 and 12 along Y-axis
I = (I 1 cos ϕ1 + I2 cos ϕ2)2 + (I2 sin ϕ2 – I1 sin ϕ1)2
If tan ϕ is +ve, then current I will lead the applied voltage V, if ,. is -ve current I will lag
Behind the applied voltage V.
Power factor of the whole circuit is given by,
Cos ϕ = I1 cos ϕ1 + I2 cos ϕ2 = X – component / I
2. Admittance method:
Admittance (denoted by symbol Y) of a circuit is defined as the reciprooal of its impedance
Y = 1 / Z = I / V or Y = r.m.s amperes / r.m.s volts
The unit of admittance is siemer1s (S). The old unit was mho (U).
As the impedance Z of a circuit has two components R and X(See Fig. 55), similarly, shown in
Fig. 56, admittance Y also has two components G (conductance-X-component) and B (susceptance Y component).
Obviously , G = Y cos ϕ = 1/ Z . R / Z = R / Z2 = R / R2 + X2
Similarly , B = Y sin ϕ = 1 / Z . X / Z = X / Z2 = X / R2 + X2
Admittance y = G2 + B2 just as Z = R2 + X2
The units of G. B and Y are in Siemens. Here, we shall consider capacitive susceptance as +ve and inductive capacitance as – ve.
Application of admittance method in solution of sing le-phase parallel circuits : Refer Fig. 57. Determine conductance and susceptance of individual branches from the relations
G = R / Z2 and B = X / Z2
Taking B as +ve if X is capacitive and a- ve if X is inductive. Let the conductances of the three branches of circuit shown in Fig 57 beG1, G2 andG3 respectively and susceptances be B
B2 and B3 respectively_ Total conductance is found by merely adding the conductances of three branches. Similarly, total susceptance is found by algebraically adding the individual susceptances of different branches.
Total conductance G = G1 + G2 + G3
and , total susceptance , B = B1 + B2 + B3
Total admittance Y = G2 + B2
Total current , I = VY
Power factor , cos ϕ = G / Y
3. Complex or Phasor algebra :
Consider the parallel circuit shown in Fig. 58 in which two impedances z1 and z2 , being in parallel, have the same potential difference across them
Now, I1 = V / Z1 and I2 = V / Z2
And current I = I1 + I2 = V / Z1 + V / Z2
= V (1 / Z1 + 1 / Z2) = V (Y + Y2 ) = VY
Where Y ( = total admittance ) = Y1 + Y2
It may be noted that the admittances are added in parallel branches, whereas impedances are added for branches in series.
It is most important to remember that admittances and impedances being complex quantities must be added in complex form.
Let us now consider the two parallel branches shown in Fig. 59, we have
Y1 = 1 / Z1 = 1/ R1 + JKL = R1 – JXL / (R1 + JKL) (R1 – JKL)
= R1 – JKL / R2 + XL2 = R1 / R1 + XL2 – J XL / R1 + XL2= G1 – JB1
Where G1 = R1 / R2 + XL2
B1 = – XL / R2 + XL2
Similarly , Y2 = 1 / Z2 = 1 / R2 – JXC = (R2 + JXC) / (R2 – JXC) (R2 + JXC)
= R2 + JXC / R2 + XC2 = R2 / R2 + XC2 + J XC / R2 + XC2 + J XC / R2 + XV2 = G2 + JB2
Total admittance Y = Y1 + Y2 = (G – JB1) + (G2 + JB1) = (G1 + G2) – J(B1 –B2) = G – JB
Y = (G1 + G2)2 + (B1 – B2)2
ϕ = tan -1 (B1 – B2 / G1 + G2)
For admittance the polar form is :
Y = Y < ϕ where ‘ is as given above
Y = G2 + B2 < tan -1 (B / G)
Total current I = V Y ;I1 = V Y1 and I2 = V Y2
If V = V < 0′ and y= Y < ϕ then j =VY = VLO’ x Y< ϕ = VY < ϕ
In general, i.e., V = V La and Y- Y 4. then
i =VY = V < a x Y < β=VY < (a + β)
Thus, it is worth noting that when vector algebra is multiplied by admittance either in com·
plex (rectangular) or polar form, the result is vector current in its proper phase relationship with respect to the voltage, irrespective of the axis to which the voltage may have been referred to.
Example 41.A resistance of 60 an inductance of 0.18 H and a capacitance of 120 µF are connected in parallel across a 100 V, 50Hz supply. Calculate:
(i) Current in each path.
(ii) Resultant current.
(iii) Phase angle between the resultant current and the supply voltage.
(iv) Power factor of the circuit.
Solution. Given: R = 60; L = 0.18 H; C = 120 µF = 120 x 10 -6 F, V = 100 volts, 50 Hz.
:. Inductance reactance, XL= 2πfL = 2π. 50 x 0.18 = 56.55 and
Capacitance reactance , XC = 1 / 2πfC = 1 / 2π x 50 x 120 x 10 -6 = 26.54
(i) Current in each path :
Current through reactance,
I1 = V / R = 100 / 60 = 1.67 A in phase with voltage V.
Current through inductance
I2 = V / XL = 100 / 56.55 = 1.77 A lagging behind voltage V by 90
Current through capacitance
I3 = V / XC = 100 / 26.53 = 3.77 A leading the voltage V by 90
The circuit and phasor diagrams are shown in Fig. 60 (c) and (b) respectively.
(ii) Resultant current, I :
Resultant current, I = (I1)2 + (I3 – I2)2 = (1.67(2 + (3.77 -1.77)2 = 6.4
(iii) Phase angle between the resultant current and the supply voltage,+ :
ϕ = tan -1 (I3 – I2 / I1) = tan -1 (2 / 1.67) = 50.14 (lead)
(iv) Power factor of the circuit, cos ϕ:
Cos ϕ cos 50.14 = 0.641 (lead). (Ans.)
Example 43. The currents in each branch of a two-branched parallel circuit are given by the Expressions I A = 7.07 sin (314t –π/4) and i8 = 21.2 sin (314t + π/ 3). The supply voltage is given by the expression v = 354 sin 314t. Derive a similar expression for the supply current and calculate the ohmic value of the components, assuming two pure components in each branch. State whether the reactive components are inductive or capacitive.
Solution. From the given expressions of currents, we find that :
• i A lags the voltage by n/4 radian or 45o and i8 leads it by n/3 radian or 60°. Hence branch A
consists of a resistance in series with a pure inductive reactance. Branch B consists of a
resistance in series with a pure capacitive reactance as shown in Fig. 62
Maximum value of current in branch A is 7.07 A and in branch B is 21.2 A.
The resultant current can be found as follows :
As seen from the vector diagram [Fig. 62 (b)],
X-component = 21.2 cos 60° + 7.07 cos 45° = 15.6 A
Y-component = 21.2 sin 60°-7.07 sin 45o = 13.36 A
Maximum value of the resuitant current
= (15. 6)2 + (13 .36)2 = 20.54 A
ϕ, = tan-1 (13.36/15.6) = 40.6° (lead)
Hence the expression for the supply current is :
i = 20.54 sin (314t + 40.6°) (Ans.)
ZA = 354 / 7.07 = 50 ; cos ϕ a = cos 45 = 0.07 sin ϕ A = sin 45 = 0.707
RA = zA cos ϕ A =50 x o.707 = 35.4 n
XL= ZA sin ϕ A =50 x0.707 = 35.4 n
ZB = 354 / 21.2 = 16.7
RB = ZB cos 60 = 16.7 x cos 60 = 8.35
XB = ZB sin 60 = 16.7 x sin 60 = 14.46
Example 44. Two impedances given by Z 1 = (1 0 + j5) and Z2 = (8 + j6) are joined in parallel and connected across a voltage of v = 200 + j0. Calculate the circuit current, its phase and the branch currents. Draw the phasor diagram. (M.S. University Baroda)
Solution. Refer Fig. 63 (a).
Branch A , Y1 = 1 / Z1 = 1 / (10 + j5)
= 1 / (10 + j5) x 10 – j5 / (10 – j5) = 10 –j5 / 100 +25 = 10-j5 / 125
= (0.08 – j0.04) siemens
Branch B , Y2 = 1 / Z2 = 1 / (8 + j6)
= 1 / (8 + j6) x (8 – j6) / (8 – j6) = 8 – j6 / 64 + 36 = 8 – j6 / 100
= (0.08 – j0.06) siemens
Y = Y1 + Y2 = (0.08 – j0.04) + (0.08 – j0.06)
= (0.16 – j0.1) siemens
1 / Z = 1 / Z1 + 1 / Z2 + 1 / Z2 = Z1 + Z2 / Z1Z2
Y = Z1 + Z2 / Z1Z2 = (10 + j5) + (8 +j6) / 10 + j5 ) (8 + j6) = (18 + j11) / 50 + j 100
Rationalising the above expression, we get
Y = (18 + j 11) (50 – j 100) / (50 + j 100) (50 – j 100) = 2000 – j 1250 / 12500
= 0.16 – j0.1 (same as before)
V = 200 < 0
I = VY = (200 + j0) (0.16 – j0.1)
= 32 – j20 = 37.74 < – 32 …polar from
It lags behind the applied voltage by 32°.
Power factor = cos 32° = 0.848. C Ans.)
I1 = VY1 = (200 + j0)(0.08 – }0.04)
= 16-j8 = 17.89 L – 26° 33′. (Ans.)
It lags behind the applied voltage by 26° 33′.
I2 = VY2 = (200 + j0)(0.08 – j0.06)
= 16-j12 = 20L- 36° 52′
It lags behind the applied voltage by 36° 52′.
The phasor diagram is shown in Fig. 63 (b).
Example 45. Fig. 64 shows a parallel circuit in which the values of the parameters are as given below :
R1 = 70 .Q (non-inductive); Coil: RC = 30 .Q, LC = 0.5 H; R2 – 100 Q; XC = 157 .Q (at 50 Hz).
(i) Determine the branch currents and the total current.
(ii) Draw the phasor diagram indicating the currents and voltages across coil and condenser.
(iii) If the A.C. source is replaced by an equivalent D.C. source, what current would be drawn by
the circuit ? (Pune University)
Solution. Given: R 1 = 70 Q; Rc = 30 n, L 0 = 0.5 H; R2 = 100 Q ;X0 = 157 n,
Applied voltage = 240 V (r.m.s.), 50 Hz
(i) The branch currents (I1‘ I2) and the total current (I) :
Resistance of the inductive branch = R1 + Rc = 70 + 30 = 100 Q
Reactance of the inductive branch = 2π f LC = 2n x 50 x 0.5 = 157 Q
Conductance of inductive branch
G = 2π f LC / (R1 + RC)2 + (2π f LC)2
= 157 / (10002 + (157)2 = – 0.00288 S (siements)
Susceptance of inductive branch ,
B1 = 2π f LC / (R1 + RC)2 + (2π f LC)2
= (100)2 + (157)2 = – 0.00453 S (Being inductive)
Conductance of capacitive branch
G2 = R2 / R2 + XC2 = 100 / (100)2 + (157)2 = 0.00288 S
Susceptance of capacitive branch,
B2 = XC / R2 + XC2 = 100 / (100)2 + (157)2 = 0.00453 S (Being inductive)
Total conductance of the circuit,
G = G1 + G2 = 0.00288 + 0.00288 = 0.00576 S
Total susceptance of the circuit,
B = B1 + B2 =- 0.00453 + 0.00453 = 0
Total admittance of the circuit,
y = G2 + B2 = (0.00576)2 + 02 = 0.00576 S
Current in inductive branch, I1 = V x Y1 = V G12 + B1
= 240 (0.00288)2 + (- 0.00453)2 = 1.288 A. (Ans.)
Phase angle ϕ1 = tan -1 (B1 / G10 = tan -1 ( -0.00453 / 0.00288) = – 57.5.
Current in capacitive branch, I2 = V x Y2 = V G2 2 + B2
= 240 (0. 00288)2 + (0. 00453)2 = 1.288 A. (Ans.)
Phase angle, ϕ1= tan -1 (B2 / G2) = tan -1 (0.00453 / 0.00288) = 57.5.
Total current, I = V x Y = 240 x 0.00576 = 1.38 A
Phase angle, ϕ = tan -1 (B / G) = tan -1 (0 / 0.00576) = 0
Voltage across condenser
= I2 Xc = 1.288 x 157 = 202.2 V lagging behind 12 by 90°. (Ans.)
Voltage across coil = I1 x (Rc)2 + (2πf Lc)2 = 1.288 (30)2 +(157)2 = 206 V
Phase angle with current I1 = tan-1
(ii) Phasor diagram :
The phasor diagram indicating the currents and voltages across coil and condenser is shown in Fig. 65.
(iii) Current drawn by the circuit when D.C. source is used :
When energised by equivalent D.C. source (i.e., 240 V D.C.), the capacitive branch will be open and current drawn by the circuit
=Current drawn by inductive branch
= V / R1 + RC = 240 / 70 + 30 = 2.4 A.
Example 46. Fig. 66 shows two impedances (18 + j24) and (15 – j30) Q connected in
parallel ; across the combination is applied a voltage of 200L53°8 Determine :
(i) kVA, k VAR and kW in each branch.
(ii) The power factor of the whole circuit.
Solution. Refer Fig. 66
Y1 = 1 / 18 + J24 = 18 –J24 / (18 + J24 ) (i18 – j24)
= 18 – j24 / 900 = (0.2 –j0.0266) S
Y2 = 1 / 15 – j30 = 15 – j30 / (15 + j30) (15 –j30) = 15 + j30 / 1125
Now, V = 200L53°8′ = 200 (cos 53°8′ + j sin 53°8′)
= 200(0.6 + j0.8) = (120 + j160) volts
I1 = V Y (120 + j160) (0.2 – J0.0266)
= 2.4 + j3.2-j3.2 + 4.26 = (6.66 + j0)
I2 = v = (120 + j160)(0.0133 + j0.0266)
= 1.6 + j3.2 + j2.13 – 4.26 = – 2.66 + j5.33 (leading)
(i) kVA, kV AR and kW in each branch :
Power calculations can be calculated by the method of conjugates.
Branch 1 (Inductive branch)
The current conjugate of (6.66 + j0) is (6.66-j0)
V f1 = (120 + 160) (6.66-j0) = 800 + 1066
KVAR = 1066 / 1000 = 1.066
The fact that it is positive merely shows that reactive VA (volt-amps) are due to lagging current
kVA = (kW)2 + (kVAR)2 = Jco.8)2 + (1.066)2 = 1.33. (Ans.)
Branch 2 (Capacitive branch) :
The current conjugate of (- 2.66 + j5.33) is(- 2.66 – j5.33)
VI 2 = (120 + j160) (- 2.66 – j5.33) = 533.6 – j1065.2
KW = 533.6 / 1000 = 0.5336 . KVAR = -1065.2 / 1000 = – 1.065 .
The negative sign merely indicates the reactive volt-amps are due to leading current.
kVA = (0.5336)2 + (-1.065)2 = 1.191. (.Ans.)
(ii) The power factor of the whole circuit, cos ϕ :
Y = Y + f; = (0.02-j0.0266) + (0.0133 + j0.0266) = 0.0333 + j0
I = V Y = (120 + j160)(0.0333 + j0) = 4 – j5.33 = 6.66 L53°8′
I =I 1 + I2 = (6.66 + j0) + (- 2.66 + j5.33) = 4-j5.33 (Same as above)
Power factor of the circuit, cos ϕ =cos 0° = 1. (Ans.)
Example 47. For the circuit of Fig. 67, calculate the current supplied by the voltage source and the voltage across the current source.
Solution. Refer Figs. 67 and 68.
Let the impedance of the current source be Z i, then
1 / Z i = 1 / 50 < 0 + 1 / 30 < 90 = 30 < 90 + 50 <0 / 1500 < 90
= 50 + j30 / 1500 <90 = 58.31 < 30 .96 / 1500 < 90
Z1 = 1500 < 90 / 58 .31 < 30 .96 = 25.72 <59 .04 = (13.23 + j22.05)
Voltage across current source,
V1 = IZ i = 2.83 L45° x 25.72 L59.04° = 72.78 L104.04° volts. (Ans.)
Current supplied by one source,
I1 = V1 / Z I + R = 72 .78 < 104 .04 / 13.25 + J22.05 + 55
= 72.78 < 104 .04 / 68.25 + J22.05
= 72.78 <104 .04 / 71.72 < 17.91 = 1.015 < 86 .13 = (0.0685 + J1.01) A
Current supplied by the other source,
I2 V2 / Z i + R = 100 < 0 / 71.72 < 17.91 = 1.394 < (-17.91) = 1.326 – j0.428
Current supplied by the voltage source,
I = I 2 – I 1 = 1.326-j0.428- 0.0685-j1.01 = 1.2575-j1.438
= 1.91 L- 48.83 amp. (Ans.)
4.6.3. Series-Parallel Circuits
Series-parallel circuits may be solved by the following methods :
1. Admittance method.
2. Symbolic method.
1. Admittance method :
In series-parallel circuits, the parallel circuit is first reduced to an equivalent series circuit and then combined with the rest of the circuit as usual.
For a parallel circuit,
Equivalent series resistance, R eq = z cos ϕ = 1 / Y . G / Y = G / Y2
Equivalent series reactance, X eq = z sin ϕ = 1/ Y . B / Y = B / Y2
2. Symbolic method :
Refer series-parallel circuit shown in Fig. 69. First calculate the equivalent impedance of
parallel branches and then add it to the series impedance to get the total impedance of the circuit. Then current flowing through the circuit is found as follows
Y2 = 1/ R2 + JX2 and Y3 = 1 / R3 – JX3
Y23 = 1/ R2 + JX2 + 1 / JX3
Z23 = 1 / Y23 ; Z1 = R1 + JX1
Z = Z23 + Z1
I = V / Z
4.6.4. Resonance in parallel circuits
In case of a series circuit consisting of R (resistance), L (inductance) and C (capacitance), resonance takes place when VL = V c i.e., when XL = Xc. In other words, resonance takes place when the power factor of the circuit approaches unity. The basic condition for resonance, i.e., power factor of the entire circuit being unity, remains the same for parallel circuits also. Thus, resonance in a parallel circuit will occur, when the power factor of the entire circuit becomes unity.
A parallel circuit consisting of an inductive coil in parallel with a capacitor is shown in Fig. 74 (a). The phasor diagram of this circuit with applied voltage as the reference phasor is shown in Fig. 74 (b). The current drawn by the inductive branch lags the applied voltage by an angle ϕ L. The current drawn by the capacitive branch leads the applied voltage by 90°.
The power factor of the current becomes unity when the total current drawn by the entire circuit is in phase with the applied voltage. This will happen only when the current drawn by the capacitive branch Ic equals the active component of the current of the inductive branch IL [Fig. 74
Hence for resonance in parallel circuit.
IC = IL sin ϕ L
Now , I L = V/ Z ; sin ϕ L = XL / Z and IC = V / XC
Hence, condition for resonance becomes
V / XC = V / Z x XL / Z or Z2 = XL x XC
Now , XL = w L, XC = 1 / w C
Z2 = WL / WC = L / C
R2 + XL2 = R2 + (2π f r L)2 = L / C
(2πf r L)2 = L / C –R2 or 2πf r = 1 / LC = R2 / L2
f r = 1 / 2π 1 / LC – R2 / L2
This is the resonant frequency and is given in Hz if R is in ohm, L is in henry and C in farad.
If R is negligible, then
f r = 1 / 2π √ LC
Current at resonance
Refer Fig. 7 4 (b). Since wattless component is zero, the circuit current is given as :
I = IL cos ϕ L = V/Z x R / Z = VR / Z2
Putting the value of Z 2 = L / C from eqn. (19), we get
I = VR / L /C = V / L/CR
Thus, the impedance offered by a resonant parallel circuit= CR.
This impedance is purely resistive and generally termed as equivalent or dynamic impedance of the circuit. As the resultant current drawn by a resonant parallel circuit is minimum, the circuit is normally called rejector circuit. Such a type of circuit is quite useful in radio work.
The phenomenon of resonance in parallel circuits is normally termed as “current resonance” whereas it is termed “voltage resonance” in series circuit.
Resonance characteristics :
Fig. 75 shows the characteristics of the parallel circuit consisting of an inductance L and capacitance C in parallel plotted against frequency, the voltage applied being constant.
Inductive susceptance, b = – 1 / XL = – 1 / WL = – 1 / 2π f L
Thus inductive susceptance is inversely proportional to the frequency and is represented by rectangular hyperbola in the fourth quadrant (because it is assumed -ve).
Capacitive susceptance, b = 1 / XC wC = 2π f C
Thus capacitive susceptance is directly proportional to the frequency and is represented by a straight line passing through the origin.
Net susceptance, B is the difference of the two susceptances and is represented by the dotted hyperbola. The net susceptance is zero at point A, hence admittance is minimum and is equal to Thus, at point A, line current is minimum. The frequency at which the total current becomes minimum is the resonance frequency fr·
Evidently, below the resonant frequency, the inductive susceptance predominates, thus making the circuit current to be lagging, whereas beyond fr capacitive susceptance predominates and the current leads the applied voltage. At resonant frequency fr’ the current is in phase with the applied voltage.
Hence at parallel resonance it is seen that :
1. The admittance of the circuit is minimum and is equal to the conductance of the circuit.
2. The current drawn is minimum.
3. The phase angle between the current and voltage is zero, the power factor is unity.
4. The resonant frequency is given by f r = 1 / 2π √LC if the resistance in the inductance and capacitance branches is negligible.
4.6.5. Comparison of series and parallel resonant circuits
4.6.6. Q-factor of a parallel circuit
It is defined as the ratio of the current circulating between its two branches to the line current drawn from the supply or simply, as the current magnification.
Q factor = 1 / R √L / C
• It may be noted that in series circuits, Q-factor givet the voltage magnification, whereas in
parallel circuits it gives the current magnification.
Bandwidth of a parallel resonant circuit :
The bandwidth of a parallel circuit is defined in the same way as that for a series circuit. This circuit also has upper and lower half-power frequencies where power dissipated is half of that at resonant frequency.
The net susceptance B, at bandwidth frequencies, equals conductance. Hence,
At f2 : B = Bc2 -BL2
At f1 : B = BL1 -Bc1
Hence Y = G2 -+ B2 = √2G and ϕ = tan-1 (1) = tan-1 (1) = 45°
However, at off-resonant frequencies, Y > G and BC= BL and phase angle is greater than zero.
Example 50. An inductive circuit of resistance 2 ohms and inductance 0. 01 His connected to a 250 V, 50 Hz supply.
(i) What capacitance placed in parallel will produce resonance ?
(ii) Determine also the total current taken from the supply and the currents in the branch
circuit. (Kerala University)
Solution. Given: R = 2 .Q ; L = 0.01 H, Supply voltage = 250 V, 50 Hz.
(i) Value of capacitance which will produce resonance, C :
Now, XL= 2π f L = 21t x 50 x 0.01 = 3.14 .Q
We know that , Z2 = L / C or L / Z2
C = 0.01 / (3.72)2 = 722.6 x 10 -6 F or 722.6 µF.
(ii) Total current and currents in the branch circuits, I, IL IC :
IR.L = V / Z = 250 / 3.72 = 67.2 A.
tan ϕl = 3.14 / 2 = 1.57 or ϕ L = tan -1 (1.57) = 57.5
Hence, current in R-L branch lags the applied voltage by 57.5°.
IC = V / XC = V / 1 / wC = wVC = 2π x 50 x 250 x722.6 x 10 -6 = 56.75
This current leads the applied voltage by 90°.
Total current, I= IR·L cos ϕ = 67.2 cos 57.5° = 36.1 A. (Ans.)