Parallel ac circuits can be solved by the following methods: 1. Phasor method 2. Admittance method.
The application of a particular method will depend upon the conditions of the problem. However, in general, that method should be used which gives quick results.
1. Phasor Method:
Consider a parallel circuit consisting of three branches of impedances Z1 (R1, L1, C1), Z2, (R2, L2, C2,) and Z3 (R3, L3, C3) respectively and connected in parallel across an alternating voltage of V volts (rms), as shown in Fig. 4.30 (a).
Since all the three branches are connected in parallel, the voltage across each branch is the same and equal to supply voltage V, but currents through them will be different.
ADVERTISEMENTS:
Branch I:
Current I1 will lead or lag behind the applied voltage if ɸ1 is + ve or – ve respectively.
ADVERTISEMENTS:
Similarly in Branch 2:
And in Branch 3:
Resultant current, I, which is phasor sum of I1, I2, and I3, can be determined either by using parallelogram law of phasors, as shown in Fig. 4.30 (b) or by resolving branch currents I1, I2, and I3 along X-axis and Y-axis and then determining the resultant of these components analytically.
ADVERTISEMENTS:
Component of resultant current I along X-axis
= Sum of components of branch currents I1, I2 and I3 along X-axis
Or I cos ɸ = I1 cos ɸ1 + I2 cos ɸ2 + I3 cos ɸ3
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Similarly component of resultant current I along Y-axis
= Sum of components of branch currents I1, I2 and I3 along Y-axis
If ɸ is + ve current I will lead the applied voltage V, if ɸ is – ve current I will lag behind the applied voltage.
2. Admittance Method:
ADVERTISEMENTS:
In the circuit shown in Fig. 4.40 (a), the branch currents:
If Z is equivalent impedance of the circuit then total current flowing through the circuit:
The reciprocal of impedance is known as admittance and is represented by English capital letter Y and expressed in Siemens (S).
Just as impedance has, two components, resistance R and reactance X similarly the admittance has two components known as conductance, G and susceptance, B respectively. The unit of conductance and susceptance is Siemens (S).
From admittance and impedance triangles (Fig. 4.40).
Application of Admittance Method in Solution of Single Phase Parallel Circuits:
Determine conductance and susceptance of individual branches from the relation- taking B as + ve if X is capacitive and as -ve if X is inductive.
Let the conductance of the three branches of circuit shown in Fig. 4.40 (a) be G1, G2 and G3 respectively and susceptances be B1, B2 and B3 respectively. Find the equivalent conductance and susceptance of the circuit which will be equal to algebraic sum of conductance and susceptances of individual branches respectively.
For the circuit shown in Fig. 4.40 (a)
Total conductance, G = G1 + G2 + G3
And total susceptance, B = B1 + B2, + B3
Total admittance of the circuit, Y= √G2 + B2
The current in the circuit may be determined from the relation:
I = V × Y having phase angle ɸ = Tan-1 B/G.
Note:
Capacitive susceptance will be taken as + ve and inductive susceptance as – ve though capacitive reactance is taken as – ve and inductive reactance as + ve. If ɸ is + ve current will lead and if ɸ is – ve current will lag behind the applied voltage.