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
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 Importance Method
 Admission 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,
I_{1} = V/Z_{1} = V/√R²_{1} +X²_{L}
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Cos_{1} = R_{1}/Z_{1} (lagging)
and_{1} = cos^{1} R_{1}/Z_{1}
Hence the current in the RL circuit_{1} Voltage vector to angle_{1} 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
I_{2} = V/Z_{2} = V/√R_{2}+ X_{C}?
Cos_{2} = R²/Z_{2} (Leading)
Therefore,_{2} = cos^{1} R²/Z_{2}
Considering the voltage vector V as the reference vector in the figure, the current I1, angle_{1} on the back and section I_{2} angle_{2} 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)
I_{X} = I_{1} cos_{1} + I_{2} cos_{2} = IcosΦ
and I_{Y} = I_{1} sin_{1} + I_{2} sin_{2} = I sinΦ
So I = (I cosΦ)² + (I sinΦ)²
phase angle of the circuit
tanΦ = I_{Y}/I_{X}
= tan1 IY/IX
Admission Method
The permittivity (Y) of the alternating current circuit is equal to the inverse of the enthalpy (Z) of the circuit.
Admittance Y = 1/Z (mho)
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.
I_{1} = V/Z_{1}
I_{2} = V/ Z_{2}
I_{3} = V/Z_{3}
vector form
I = I_{1} + I_{2}
Total impedance Z of the circuit
1/Z = 1/Z_{1} + 1/Z_{2} + 1/Z_{3}
or Y = Y_{1} + Y_{2} + Y_{3}
and conductivity G = G_{1} + G_{2} + G_{3}
and compatibility B = B_{1} + B_{2} + B_{3}
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 Xaxis, it becomes jP on the +Yaxis 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 = (A_{1})² + (A_{2})²
and tanΦ = A_{2}/A_{1}
and P = (P_{1})² + (P_{2})²
and tanΦ_{1} = – P_{2} /P_{1}
The two components of impedance in an alternating current circuit can be represented as follows.
Z = R + jX_{L }
Z = (R)² + (X_{L})²
Z_{1} = R_{1} + jX_{C }
Z = (R)_{1})² + (X_{L})²
A complex impedance Z = R² + (X_{L} – X_{C})² can be done by the method as follows.
in the picture,
Z = R + jX_{L} – jX_{C}
= R + j(X_{L} – X_{C},
Current of the circuit I = V/Z
= (V + j 0)/(R + jX_{L} – jX_{C},
= I_{1} + jI_{2}
and I = (I_{1})² + (I_{2})²
If the voltage and current in a circuit are as follows –
V = V_{1} + jV_{2}
I = I_{1} + jI_{2}
Then, the different powers of the circuit can be determined by the operator as follows
Real Power
P_{T} = V_{1}I_{1} + V_{2}I_{2}
Reactive Power
P_{R} = (V_{1}I_{2} – V_{2}I_{1},
dwelling power
P_{A} = P²_{T} + P²_{R}
and power factor
CosΦ = P_{T}/P_{A}
Solution of parallel circuit by j – Method
I_{1} = V/Z_{1} = (V + j0)/(R)_{1} + jX_{1} ,
= A_{1} + jB_{1}
I_{2} = (V + j0 )/ (R_{2} – jX_{2},
= A_{2} – jB_{2} (say)
So, I = (i)_{1})² + (i_{2})²
Power dissipated in the circuit = V × conjugate of current
= (V + j0) (i)_{1} + jI_{2},
= Vi_{1} + jVi_{2}
Trigonometrical Method
in the picture,
A_{1} = Acos?
A_{2} = Acosθ
A_{1} and A_{2} can be written by j method as follows.
A_{1} = Acosθ
A_{2} = jAcosθ
A = Acosθ + jAsinθ
Similarly, if vector A is specified on the xaxis, then
A = A cosθ jA sinθ
Exponential form
Euler’s equation is used in this method. According to the particle at this time,
e^{j} = cosO + jsinθ
Ae^{j} = 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 Xaxis. ? 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 < (θ_{1}) = (A cos + B cos_{1}) + j(A sinθ – B sinθ_{1},
A ? / B < (?_{1}) = (A/B) ?, ?_{1},
Also read –
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