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5. What are alternating current parallel circuits? (What is AC Parallel circuit)

# What are alternating current parallel circuits? (What is AC Parallel circuit)

Hello friends In this article we will know what are AC parallel circuits? and learn different methods of solving different circuits. And will learn about many facts related to alternating current parallel circuit.

## AC parallel circuit

Following methods of solving AC parallel circuit are-

• Importance Method
• Complex Notation Method

## Importance Method

In this method, after finding the impedance and current of each branch of the circuit, other parameters of the circuit are determined by vector diagram.

Example: In the circuit shown in the figure, two branches one RL and the other RC branch are connected in parallel. Both the combinations get the same voltage V.
Therefore

In branch RL,

I1 = V/Z1 = V/√R²1 +X²L

Cos1 = R1/Z1 (lagging)

and1 = cos-1 R1/Z1

Hence the current in the RL circuit1 Voltage vector to angle1 But will be backward.

In RC branch,

The reactance of this combination will be capacitive. so it flows stream I1 will be leading from the voltage vector

I2 = V/Z2 = V/√R2+ XC?

Therefore,2 = cos-1 R²/Z2

Considering the voltage vector V as the reference vector in the figure, the current I1, angle1 on the back and section I2 angle2 drawn on the foreground.
If the total current in the circuit is I and the power factor is , from Fig. 5.3(a) and 5.4(b)

IX = I1 cos1 + I2 cos2 = IcosΦ

and IY = I1 sin1 + I2 sin2 = I sinΦ

So I = (I cosΦ)² + (I sinΦ)²

phase angle of the circuit

tanΦ = IY/IX

= tan-1 IY/IX

The permittivity (Y) of the alternating current circuit is equal to the inverse of the enthalpy (Z) of the circuit.
The permittivity Y of the circuit is a vector quantity. Its two components are Conductivity (G) and Susceptance ‘B’.

Susceptance is also of two types, capacitive and inductive. Capacitive B is positive and inductive B is negative.

The Admittance triangle is displayed in the figure.

Condutance ‘G’ = Y cosΦ = (1/Z)(R/Z) = R/Z²

= R/(R² + X²)

Susceptance ‘B’ = Y sinΦ = (1/Z).(X/Z) = X/Z²
and X/R² + X²

Y = G² + B²

Current in the circuit I = V/Z = VY
and power multiplication cosΦ = G/Y
Power dissipated in the circuit P = VI cosΦ

For example, if an alternating current parallel circuit has three branches then it can be solved as follows.

I1 = V/Z1
I2 = V/ Z2
I3 = V/Z3
vector form
I = I1 + I2

Total impedance Z of the circuit

1/Z = 1/Z1 + 1/Z2 + 1/Z3
or Y = Y1 + Y2 + Y3
and conductivity G = G1 + G2 + G3
and compatibility B = B1 + B2 + B3
Y = G² + B²
and I = VY

## Complex Notation Method

There are four types of complex notation:

• Symbolic form
• Trigonometrical form
• Exponential form

### Symbolic form

In this method j operator is used to solve the alternating current circuit. By operating j on any vector, it rotates by 90°.
For example, when the vector P is operated on the positive direction of the X-axis, it becomes jP on the +Y-axis and j²P on the -X axis, j³P on the -Y axis and again j⁴P on the +X. So according to the figure,

j²P = -P
j² = -1
j = -1
j = j². j
= – j
j⁴ = j² × j²
= (-1) × (-1)
= +1

Similarly, the two components of a vector acting at angle O can be represented as follows.

A = A1 + jA2

Similarly P = -P1 + jP2

According to the above equation and Fig. 5.38 and 5.39, the amplitude of vectors A and P can be found as follows.

|A| = (A1)² + (A2
and tanΦ = A2/A1
and |P| = (-P1)² + (-P2
and tanΦ1 = – P2 /-P1

The two components of impedance in an alternating current circuit can be represented as follows.

Z = R + jXL

Z = (R)² + (XL
Z1 = R1 + jXC

Z = (R)1)² + (XL

A complex impedance Z = R² + (XL – XC)² can be done by the method as follows.

in the picture,

Z = R + jXL – jXC
= R + j(XL – XC,

Current of the circuit I = V/Z
= (V + j 0)/(R + jXL – jXC,
= I1 + jI2

and |I| = (I1)² + (I2

If the voltage and current in a circuit are as follows –

V = V1 + jV2
I = I1 + jI2

Then, the different powers of the circuit can be determined by the operator as follows-

Real Power-
PT = V1I1 + V2I2
Reactive Power-
PR = (V1I2 – V2I1,
dwelling power
PA = P²T + P²R
and power factor

CosΦ = PT/PA

### Solution of parallel circuit by j – Method

I1 = V/Z1 = (V + j0)/(R)1 + jX1 ,

= A1 + jB1
I2 = (V + j0 )/ (R2 – jX2,
= A2 – jB2 (say)

So, |I| = (i)1)² + (i2

Power dissipated in the circuit = V × conjugate of current

= (V + j0) (i)1 + jI2,

= Vi1 + jVi2

### Trigonometrical Method

in the picture,

A1 = Acos?
A2 = Acosθ

A1 and A2 can be written by j method as follows.

A1 = Acosθ
A2 = jAcosθ
A = Acosθ + jAsinθ

Similarly, if vector A is specified on the x-axis, then

A = A cosθ- jA sinθ

### Exponential form

Euler’s equation is used in this method. According to the particle at this time,

ej = cosO + jsinθ
Aej = Acosθ + jAsinθ
Ae-jθ = Acosθ – jA sinθ

### Polar form

In this method the vector is represented by its magnitude and phase angle. for example ? There is a vector whose result is R and it is from the X-axis. ? angle is specified.

from the picture,

R ? = Rcos?+ jRsin?
and P ?1 = P cos?1 – jP sin?1

Two vector A in polar form ? and B ?1 Mathematical operations can be done as follows.

(A ?) × (B ?1) = A × B { ? + (-1,
A < + B < (-θ1) = (A cos + B cos1) + j(A sinθ – B sinθ1,
A ? / B < (-?1) = (A/B) ?, ?1,