# A.C. series circuits

Under this heading we shall discuss R-L, R-C and R-L-C series circuits..

4.5.1. R-L circuit (Resistance and inductance in series)

Fig. 33 (a) shows a pure resistance Rand a pure inductive coil of inductance L connected in series. Such a circuit is known as R-L circuit (usually met a cross in practice).

Fig. 33. R-L circuit (Resistance and inductance in series).

Let       V = R.M.S. value of the applied voltage,

I = R.M.S. value of the resultant current,

VR = IR = Voltage drop across R(in phase with n, and

VL = IXL= Voltage drop across L (coil), ahead of I by 90°.

The voltage drop VR and VL and shown in voltage triangle OAB in Fig. 33 (b), /being taken as the reference vector in the phasor diagram. Vector OA represents ohmic drop VR and AB represents inductive drop VL. Vector OB represents the applied voltage V which is the vector sum of the two (i.e:, VR and VL).

where Z = R2 + XL2 (total opposition offered to the flow of alternating current by R-L series circuit) is known as impedance of the circuit.

As seen from the “impedance triangle” ABC [Fig. 33 (c)],

z2 = R2 +XL2

i.e., (Impedance)2 = (Resistance)2 + (Inductive reactance?

From Fig. 33 (b) it is evident that. voltage V leads the current by an angle 4> such that,

tan ϕ = VL / VR = LX / IR = XL / R = wL / R = Inductive reactance / Resistance

ϕ = tan -1 ( XL / R)

The same is illustrated graphically in Fig. 33 (d).

In other words I lags V by an angle ϕ

Power factor, cos ϕ = R / Z [From Fig. 33 (c)]

Thus, if the applied voltage is given by v = V max sin wt , then current equation is given as,

i = Imax sin (wt – ϕ)

I max = V max / Z

In the Fig. 33 (e), I has been shown resolved into two components, I cos 4> along V and I sin 4 in quadrature (i.e., perpendicular) with V.

Mean power consumed by the circuit

= V x I cos 4> (i.e., component of I which is in phase with V

i.e.,                              P = VI cos 4> (= r.m.s. voltage x r.m.s. current x cos 4 >)

The term ‘cos ϕ is called the power factor ( = R / Z) of the circuit

It may noted that :

– In A.C. circuit the product of r.m.s. volts and r.m.s. amperes gives volt-amperes (i.e., VA)

and not true power in watts. True power (W) = volt-amperes (VA) x power factor

or                     Watts = VA (Apparent power) x cos 4>

-The power consumed is due to ohmic resistance only since pure inductance consumes no

power.

i.e,                   p = VI cos ϕ = VI x R / Z = V / Z = I2R , watts

This shows that power is actually consumed in resistance only; the inductor does not consume any power.

The power consumed in R-L circuit is shown graphically in Fig. 33 (/).

Thus in R-L circuit we have :

1. Impedance, Z = R2 + XL2 (where XL= wL = 2πx f L)

2. Current, I = V / Z

3. Power factor, cos ϕ = cos -1 R / Z ( = True power / Apparent power = W / VA)

(or angle of lag, ϕ = cos-1 (R / Z)]

4. Power consumed, P = VI cos ϕ ( = IZ x I x R / Z = I2 R)

Symbolic Notation :

Z=R + j XL

The numerical value of impedance vector = R2 + XL 2

The phase angle with the reference axis, ϕ = tan-1 (XL / R).

In polar form : Z = Z < ϕ

Apparent, Active (True or real) and Reactive Power :

Every circuit current has two components : (i) Active component and (ii) Reactive component.

“Active component” consumes power in the circuit while “reactive component” is responsible for the field which lags or leads the main current from the voltage.

In Fig. 34. active component is /active= I cos ϕ, and reactive component is /reactive = I sin ϕ

Refer Fig. 35.

(i) Apparent power (8). It is given by the product of r.m.s. values of applied voltage and

circuit current.

S =VI= (I x Z) .I= I2Z volt-amperes (VA)

(ii) Active or true or real power (P or W). It is the power which is actually dissipated in the

circuit resistance.

P = FR = VI cos ϕ watts

(iii) Reactive power (Q). A pure inductor and a pure capacitor do not consume any power, since in a half cycle what so ever power is received from the source by these components the same is returned to the source. This power which flows back and forth (i.e., in both directions in the circuit) or reacts upon itself is called “reactive power.”

It may be noted that the current in phase with the voltage produces active or true or real power while the current 90° out of phase with the voltage contributes to reactive power.

In a R-L circuit, reactive power which is the power developed in the inductive reactance of the circuit, is given as :

Q = I2 XL = I2 Z sin ϕ = I .(IZ) sin ϕ

= VI sin ϕ volt – amperes – reactive (VAR)

These three powers are shown in Fig. 35

Relation between VA, Wand VR

W = VA cos ϕ

VAR = VA sin ϕ

VA = W / cos ϕ

VA = VAR / sin ϕ

Power factor (p.f)              = W / VA = True power / Apparent power

The power factor depends on the reactive power component. If it is made equal to the active power component, the power factor becomes unity.

Example 20. A coil takes 2.5 amps. when connected across 200 volt 50 Hz mains. The power consumed by the coil is found to be 400 watts. Find the inductance and the power factor of the coil.

Solution. Current taken by the coil, I= 2.5 A

Applied voltage,         V = 200 volts

Power consumed,        P = 400 W

We know that                         P =VI cos q,

400 = 200 x 2.5 x cos ϕ or cos ϕ = 400 / 200 x 2.5 = 0.6

Hence power factor of coil is 0.8. (Ans.)

Impedance of the coil,                        Z = V / I = 200 / 2.5 = 90

Also                                         XL / Z = sin ϕ

XL = Z sin ϕ

XL = 2πfL

L = XL / 2πfL = 48 / 2π x 50 = 0.1529 H (henry).

Example 21. A 100 V, 80 W lamp is to be operated on 230 volts, 50 Hz A. C. supply. Calculate the inductance of the choke required to be connected in series with lamp for its operation. The lamp can be taken as equivalent to a non inductive resistance.

Solution. Current through the lamp when connected across 100 V supply,

I = W / V = 80 / 100 = 0.8 A

Resistance of the lamp, R = V / I = 100 / 0.8 = 125

If a choke of inductance L henry is connected in series with the lamp to operate it on 230 V, the current through the choke will also be 0.8 A.

The impedance of the circuit when choke is connected in series with the lamp,

Z = V / I = 230 / 0.8 = 287.5

Reactance of choke coil, XL =

XL = 2πFL

or                                             L = XL / 2πFL = 258.5 / 2π x 50 = 0.825 H

Hence inductance of choke coil, L = 0.825 H. (Ans.)

Example 22. A coil has a resistance of 5 Q and an inductance of 31.8 mH. Calculate the current taken by the coil and power factor when connected to 200 V, 50 Hz supply.

Draw the vector diagram.

If a non-inductive resistance of 10 n is then connected in series with coil, calculate the new value of current and its power factor.

Solution.                              R = 5 n

L = 31.8 m H or 0.0318 H

XL = 2rtfL

= 2rt x 50 x 0.0318 = 10 Q

Impedance of the coil,

Current taken by the cod, I = V / Z = 200 / 11.18 = 0.4475

Fig. 36 (b) shows the vector diagram.

When non-inductive resistance of 10 n is connected in series with the coil:

Total resistance in the circuit, R’ = 5 + 10 = 15 Q

Reactance in the circuit, XL’ =XL = 10 Q

Lmpedance of the circuit,       Z = R2 + X L2 = 152 + 102 = 18 Q

Current through the circuit     I = V / Z = 200 / 18 = 11.11 A

Power factor of the circuit, cos ϕ = R / Z = 15 / 18 = 0.833

Example 23.A current of5Aflows through a non-inductive resistance in series with a choking coil when supplied at 250 V, 50 Hz. If the voltage across the resistance is 125 V and across the coil200 V, calculate :

(i) Impedance, reactance and resistance of the coil,

(ii) The power absorbed by the coil,

(iii) The total power.

Draw the vector diagram.                                                       (Elect. Engg. Madras Univ.)

Solution. Non-inductive resistance connected in series with coil = 5 = 25 n

Refer Fig. 37 (b).

BC2 + CD2 = (200)2 = 40000

(125 + BC)2 + CD2 = (250)2 = 62500

Subtracting eqn. (i) from eqn. (ii), we get

(125 + BC)2– BC2 = 62500 – 40000 = 22500

15625 + BC2 + 250BC – BC2 = 22500

BC = 27.5V; CD= (200)2 – (27.5)2 = 198.1 V

(i) Coil impedance,                Z = 200 / 5 = 40

VR = IR = BC = 27.5             or 5R = 27.5

R = 27.5 / 5 = 5.5

Also                    VL \ IXL = CD = 198.1

XL = 198.1 / 5 = 39.62

(ii) Power absorbed by the coil,

P = I2R = 52 x 5.5 = 137.5 W. (Ans.)

(iii) The total power               =VI cos ϕ = 250 x 5 x AC / AD

= 250 x 5 x (125 + 27.5) / 250 = 762.5 W

The vector diagram is shown in Fig. 37 (b). (Ans.)

Example 24. An iron-cored coil has a D.C. resistance of 6 ohms. When it is connected

to 230 V, 50 Hz mains, the current taken is 3.5 A at a power factor of 0.5. Determine :

(i) Effective resistance of the coil .

(ii) Inductance of the coil.

(iii) Resistance which represents the effect of the iron loss.

Solution. Given : D.C. resistance (True resistance), R = 6 Q ; supply voltage = 230 V, F = 50 Hz, I = 3.5 A; p.f. = 0.5.

(i) Effective resistance of the coil, Re :

Total power consumed by the iron-cored choke coil,

P = Power loss in ohmic resistance + Iron loss in core = I2R +P1

P / I2 = R + P1 / I2 , where P / I2  is known as effective resistance of the coil

Effective resistance, Re = P / I2 = VI cos ϕ / I2 = 230 x 3.5 x 0.5 / (3.5)2 = 32.86

(ii) Inductance of the coil, L :

Impedance of the coil,                        Z = V / I = 230 / 3.5 = 65.7

Inductive reactance of the coil,

XL = Z2 – R2 = (65.7)2 – 932.86)2 = 56

L = XL / 2Πf = 56.9 / 2π x  50 = 0.1811 H

(iii) Resistance representing iron loss :

Since                           P / I2 = R + Pi / I2

Effective resistance, Re = True resistance + Resistance representing iron loss

32.86 = 6 + Resistance representing iron loss

. . Resistance representing iron loss = 32.86 – 6 = 26.86 Q. (Ans.)

Example 25. Three coils connected in series across a 100 V, 50 Hz supply have the following parameters :

R1 = 18 n, L 1 = 0.012 H; R2 = 12 .Q, L 2 = 0.036 H; R3 = 3.6 .Q, L3 = 0.072 H

Determine the potential drop and phase angle for each coil.

Solution. Fig. 38. shows the circuit diagram.

Total resistance in the circuit, R = R1 + R2 + R3 = 18 + 12 + 3.6 = 33.6 Q

Total inductance in the circuit, L = L 1 + L2 + L 3 = 0.012 + 0.036 + 0.072 = 0.12 H

Impedance of coil-1,

Current through the circuit, I= I = V / Z = 100 / 50.5 = 1.98 A

Potential drop across coil-1, V1 = IZ1 = 1.98 x 18.39 = 36.41 V (Ans.)

Potential drop across coil-2, V2 = IZ2 = 1.98 x 16.49 = 32.65 V (Ans.)

Potential drop across coil-3, V3 = IZ3 = 1.98 x 22.90 = 45.34 V (Ans.)

Phase angle of coil- 1, ϕ1′ cos-1 (R/Z1) = cos-1 (18/18.39) = 11.82°. (Ans.)

Phase angle of coil- 2, ϕ2 = cos-1 (R2/Z2) = cos-1( 12/16.49) = 43.3°. (Ans.)

Phase angle of coil- 3, ϕ3 = cos-1 (R3/Z3) = cos-1(3.6/22.90) = 80.96°. (Ans.)

4.5.2. R-C circuit (Resistance and capacitance in series)

Fig. 41. (a) shows a pure resistance R (ohms) and a pure capacitor of capacitance C (farads) connected in series. Such a circuit is known as R-C circuit.

Let, V = R.M.S. value of the applied voltage,

I= R.M.S. value of the resultant current,

VR = IR = Voltage drop across R (in phase with I) and

V c = IX c =Voltage drop across C, lagging I by 90°.

Voltage drops VR and VL are shown in voltage triangle OAB in Fig. 41 (b)/being taken as the reference vector in the phasor diagram. Vector OA represents ohmic drop VR and AB represents the capacitive drop Vc. Vector OB represents the applied voltage V, which is the vector sum of the VR and Vc)

where Z = R2 + Xc2 (total opposition offered to the flow of alternating current by R-C series circuit) is known as the impedance of the circuit.

As seen from the “impedance triangle” ABC [Fig. 41 (c)],

z2 = R2 +Xc2

i.e.,      (lmpedance)2 = (Resistance)2 + (Capacitive reactance)2

From Fig. 41 (b) it is evident that I leads the voltage V by an angle ϕ such that,

tan ϕ = VC / VR = IXC / IR = XC / R = (1 / Wc) / R = Capacitive reactance / Resistance

ϕ = tan -1 (XC / R)

The same is illustrated graphically in Fig. 41 (d).

In other words I leads VR by an angle\$.

Power factor, cos ϕ = R / Z [From Fig. 41 (c)]

Power. Refer Fig. 41 (d ),

Instantaneous power, p = vi = V max sin rot x /max sin (wt + ϕ)

= V max Imax / 2 x 2 sin (wt + ϕ ) sin wt

= V max / 2 x Imax / 2 [ cos ϕ – cos (2 wt + ϕ)]

Average power consumed in the circuit over a complete cycle,

P = Average of V max / 2 . I ma /2 cos ϕ – Average of V max / 2 . Imax / 2 cos (2wt + ϕ)

or                                 p = V max / 2 . I max  2 cos ϕ – zero

p = v r.m.s. X I r.m.s. cos ϕ =VI cos <P

where cos ϕ is the power factor of the circuit

Alternatively,              p = VI cos ϕ =I Z x I x R / Z = I2R

This shows that power is actually consumed in resistance only ; the capacitor does not consume any power.

Thus in R-C circuit, we have :

1. Impedance , z = R2 + XC2 (where XC  = 1/ wc = 1/ 2πf , c being in farad)

2. current , I = V / Z

3. Power factor , cos ϕ = R /Z ( = True power / Apparent power = W / VA)

[ or angle of lead , ϕ = cos -1 (R/Z]

4. Power consumed, P = VI cos ϕ (= I2R).

4.5.3. R-L-C circuit (Resistance, inductance and capacitance in series)

Fig. 45 shows a R-L-C circuit.

Important formulae :

1. Impedance,             Z = R2 + (XL – XC)2

2. Current                    I = V / Z

3. Power factor , cos ϕ = R / Z

4. Power consumed = VI cos ϕ (= I2 R)

Resonance in R-L-C circuits

Refer Fig. 45 (a).

The frequency of the voltage which gives the maximum value of the current in the circuit is called resonant frequency, and the circuit is said to be resonant.

At resonance,              XL = XC

2πF r L = 1 / 2π f r C

F r = 1 / 2π √ LC

where F r = Resonance frequency in Hz; L = Inductance in henry ; and C = Capacitance in farad.

Fig. 46 shows variation of XL, Xc and X (total reactance = XL – Xc) with variation of frequency f.

Fig. 4 7 shows the variation of current en with frequency (f).

At series resonance, it is seen that :

1. Net reactance of the circuit is zero i.e., XL – Xc = 0 or X= 0.

2. The impedance of the circuit is minimum and equal to the resistance (R) of the circuit

(i.e., I = V / R) Consequently circuit admittance is maximum.

3. The current drawn is maximum (i.e., I r = I max).

4. The phase angle between the current and voltage is zero; the power factor is unity.

5. The resonant frequency is given by f r =1 / 2n.√LC ; if the frequency is below the resonant frequency the net reactance in the circuit is capacitive and if the frequency is above the resonant frequency, the net reactance in the circuit is inductive.

6. Although VL = Vc, yet v coil is greater than Vc because of its resistance.

Half power frequencies, Bandwidth and Selectivity

Half power (cut-off) frequencies :

The half power frequencies are those frequencies at which the power dissipation in the circuit is half of the power dissipation at resonant frequency f r They are the corresponding frequencies {1 and {2 at the value of current/= I/J2 ; where I r is the current at resonance in R-L-C series circuit (Refer Fig. 48).

Hence power, P r drawn by the circuit at the resonance is

P r = I r2 R

Power in the circuit at F1 = (I / 2)2 R = 1/2 I r2 R

Also ,                           f1 = f r – R / 4πL

F2 = F r – R / 4πL

F2 – f1 = R / 2πL

Bandwidth and Selectivity :

The difference ({2 – { 1) is called the bandwidth (B h) of the resonant network.

The ratio of the bandwidth to the resonance frequency is defined as the selectivity of the circuit.

When frequency is varied in R-L-C circuit, the selectivity becomes

*(f2 – f1) / f r = 1/ Q

Where Q r is the quality factor of the resonant circuit.

*Relation between bandwidth and quality factor in series resonant conditions :

A series R-L-C circuit is considered. The resonant frequency and angular frequency are expressed by f r and w r respectively. In the above circuit, the current([) can be described as follows :

where V, R, L and C are the source voltage, resistance, inductance and capacitance of the circuit respectively.

The current, at a power, half of the maximum power developed at resonant frequency, is

where I, is the series resonant current i.e. R .

According to the definition of bandwidth,

wL – 1 / wC  = R

According to Fig. 48

w2 L – 1 / w2 L = R

Adding equations (i) and (ii), we get

(w1 + w2) L – 1/ C (1 / w1 + 1 / w2) =0

(w1 + w2) L – 1/ C (w1 + w2 / w1w2) = 0

w1 + w2 = 0 , w1w2 = 1/ LC

Again w r 2 = 1 / LC , [w r, is the angular frequency at resonant condition.]

Subtracting equations (i) and (ii), we have

L (w2 –w1) + 1/ C (w2 – w1 /w2w1) = 2R

(w2 – w1) + 1 / LC (w2 – w1 / w2w1) = 2R / L

(w2 –w1) + w1w2 . (w2 – w1) / w1w2 = 2R / L

(w2 – w1) = R / L

w2 – w1 / w0 = R / w0L

w2 – w1 / w0 / Q

F2 – F1 = F r / Q

Q-factor of a resonant series circuit :

The Q-factor of an R -L-C series circuit can be defined in the following different ways :

(i) Q factor is defined as the voltage magnification in the circuit at the time of resonance.

Since at resonance current is maximum i.e., I r = V/ R , the voltage across either coil or capacity or = I r X L r or I Rx Cr and supply voltage, V = I r R.

Voltage magnification = VL r / V = I r X Lr = X Lr / R= w Rl / R = Reactance / Resistance

V cr / V = I r x C r / R = X cr / R = Reactance / Resistance = 1 / w r CR

Q – factor = w r L / R = w r L / R = 2πF r L / R = Reactance / Resistance =1 / w r CR

where 9 is the circuit power factor angle of the coil.

(At resonance, circuit phase angle 9 = 0, and Q = tan 9 = 0)

(ii) The Q-factor may also be defined as under :

Q – factor = 2π maximum stored energy / energy dissipated per cycle

But resonant frequency, f r = 1 / 2πf r LC      or    2πf r – 1 / √ LC

Putting this value in eqn. 8(a), we get

Q – Factor  = 2πf r L / R = L / R x 1 / LC = 1/ R √ L /C

In series resonance, higher quality factor i.e., Q-factor means higher voltage magnification as well as higher selectivity of the tuning coil.

Example 32. A resistance 12 n, an inductance of 0.15 Hand a capacitance of 100 1J.F are connected in series across a 100 V, 50 Hz supply. Calculate :

(i) The current.

(ii) The phase difference between current and the supply voltage.

(iii) Power consumed.

Draw the vector diagram of supply voltage and the line current.
Solution. Given:             R = 12 n, L = 0.15 H or XL = 2rr{L

=  21t x 50 x 0.15 =  47.1 Q

c = 100 µF = 100 x 10-6F

XC = 1 / 2π fC  = 1 / 2π x 50 x 100 x 10 -6 = 31.8

(i) The current , I :       Z = R2 + (XL – XC)2

= 12 2 + (47.1 – 31.8)2 = 19.43

Current                                   I = V / Z = 100 / 19.43 = 5.15 A.

(ii) Phase difference, ϕ :

ϕ = cos -1 R / Z [ or tan -1 XL – XC / R]

= cos -1 12/19.43 [ or tan -1 15.3 / 12.0] = 52 (lag)

Hence current lags supply voltage by 52°. (Ans.)

(iii) Power consumed, P:

P=VI cos ϕ

= 100 x 5.15 x cos 52° = 371.1 W. (Ans.)

Fig. 49 (a), (b) show the circuit and vector/phasor diagrams respectively.

Example 33. For the circuit shown in Fig. 50 find the values of (i) current I, (ii) V1 and V2 and (iii) p.f.

Draw the vector diagram.

Solution. Refer Fig. 50

R = 10 + 20 = 30

L = 0.5 + 0.1 = 0.15 H

XL =2π FL = 2π x 50 x 0.15 = 47.1

XC = 1/ 2π FC = 1 / 2π x 50 x 10 -6 x 50

X = 47.1 – 63.7 = – 16.6

Z = R2 + X2 = (30)2 + (-16.6)2 = 34.3

(i)                                             I = V / Z = 200 / 34.3 = 5.83 A.

(ii)                                            XL1 = 2π x 50 x 0.05 = 15.7

Z1 = 102 + 15.7 2 = 18.6

V1 = IZ 1 (10 / 18.6 ) = 57 .5

ϕ1 = cos -1 (10 / 18.6) = 57.5

X L2 = 2π x 50 x 0.1 = 31 .4

X = 31.4 – 63 .7 = -32.3

Z2 = 202 + (-32.3)2 = 38

V2 = IZ2 = 5.83 x 38 = 221.5 V

ϕ2 = cos -1 (20 / 38) = 58.2

(iii)       combined p.f        = cos ϕ = R / Z = 30 / 34.3 = 0.875

Vector diagram is shown in Fig. 50 (b).

Example 34. For the circuit shown in Fig. 51. Calculate:

(i) Current;                                                      (ii) Voltage drops VpV2 and V3 ;

(iii) Power absorbed by each importance ;    (iv) Total power absorbed by the circuit.

Take voltage vector along the reference axis.

Solution.      Z1 = (8 + j6) n ; Z2 = (12 – j16) n ; Z3 = (8 + JO)

z = Z1 + Z2 + Z3 = (8 + j6) + (12 – }16) + (8 + JO) = (28-j10) n

Taking             v = v L0° = 200L0° = (200 + j0)

I = V / Z = 100 /(28 – j 10) = 200 (28 + j10) / (28 – j 10) (28 + j 10) = 100 (28 + j 10) / (28 )2 + (10)2

= 200 (28 + j 10 ) / 884 = 3.17 + j 1.13

(i) Magnitude of current = (3 .17)2 + (1.13)2 = 3.36 A. (Ans.)

(ii)                                V1 = IZ1 = (3.17 + jl.13)(8 + j6)

= 3.17 x 8 + 3.17 xj6 + 8 xjl.13 + jl.13 xj6

= 25.36 + j19.02 + j9.04 – 6.78 = 18.58 + j28.06. (Ans.)

V2 = IZ2 = (3.17 + jl.13)(12 – j16)

= 38.04-j50.72 + j13.56 + 18.08 = 56.12-j37.16. (Ans.)

V3 = IZ3 = (3.17 + jl.13)(8 + j0) = 25.36 + j9.04. (Ans.)

[V = V1 + V2 + V3 = (18.58 + j28.06) + (56.12 – j37.16)

+ (25.36 + j9.04) = 100 + j0 (check)]

Example 35. Fig. 52 shows a circuit connected to a 230 V, 50 Hz supply. Determine the following

(i) Current drawn                    (ii) Voltages V1 and V2

(iii) Power factor.

Draw also the phasor diagram.

Solution. Refer Fig. 52.

Given:         R1 = 18 , L 1 = 0.048 H; R2 = 12  L 2 = 0.012 H; C = 120 µ = 120 x 10-6 F.

XL1 = 2nfL1 = 2n x.50 x 0.048 = 15.08  XL2 = 2πFL2 = 2n x 50 x 0.012 = 3.77 Q

XC2 = 1/ 2Πfc = 1/ 2π x 50 x120 x 10 -6 = 26.52

Impedance,            Z1 = R1 + J X Ll = 18 + j15.08 = 23.48 L39.96°

Impedance,             Z2 = R2 + jXL2 – jX c = 12 + }3. 77 – }26.53 = 12 – j22. 76 = 25.7< L- 62.2°

Total impedance,        Z =Z1 + Z2 = (18 +j15.08) + (12-j22. 76) = 30j

Current drawn, I :

Taking supply voltage as reference vector, V = V < 00 = 230< 0°.

Current            I  = V / Z = 230 < 0° / 30.97 < – 14.36 A

Voltages VI and v2: Voltage,V2 =Iz1 = 7.43Ll4.36° x 23.48L39.96°

= 7.43 x 23.48 L(14.36° + 39.96°)

= 174.46 L54.32° V. (Ans.)

Voltage = iz2 = 7.43L14.36° x 25.73L-62.2°

= 7.43 x 25.73 L(14.36° – 62.2°)

= 191.2 L- 47.84°. (Ans.)

Phase angle between supply voltage and current i.e; V and I , ϕ = 14.36 (lead)

Power factor, cos ϕ :

cos ϕ = cos (14.36°)