__AC Parallel Circuits__

__AC Parallel Circuits__

__INTRODUCTION__

__INTRODUCTION__

There are more applications in alternating-current work for parallel circuits than for series circuits. Nearly all commercial, industrial, and residential power circuits are connected in parallel. The voltage across any branch circuit in a parallel arrangement is the same as the line voltage. This is true if the voltage drop in the line wires is neglected. Recall from earlier studies that the total (line) current for an ac parallel circuit may not be equal to the arithmetic sum of the current in each of the branch circuits. The currents in the branch circuits may be out of phase. Thus, the total (line) current must be calculated using vectors. The approximate value of the line current can also be found by drawing a vector diagram accurately to scale.

This unit develops methods of analyzing ac parallel circuits. These methods are based on concepts learned in the study of series circuits.

**The voltage is constant across each branch of a parallel circuit. Thus, all angles are measured with respect to the voltage vector. This means that the voltage vector is a reference vector.**

**PARALLEL CIRCUIT WITH RESISTIVE LOAD**

The first parallel circuit to be studied has a noninductive resistive load in each branch (Figure 8–1). All of the branch currents for this circuit are in phase with the line voltage. In this case, both the vector sum and the arithmetic sum of the branch currents give the total line current.

This parallel circuit is connected to a 120-V, 60-Hz source. The calculations are the same for an ac parallel circuit with noninductive resistance loads only and for a direct- current parallel circuit.

**PROBLEM 1**

**Statement of the Problem**

For the parallel circuit shown in Figure 8–1, find the following quantities:

1. The combined resistance

2. The current taken by each branch

3. The line current

4. The total power taken by the parallel circuit

5. The power factor and the phase angle

6. The vector diagram

**Solution**

1. The combined resistance is obtained by using the reciprocal resistance formula (discussed in *Direct Current Fundamentals*):

2. The current taken by each branch is given by Ohm’s law:

3. The total current can be found by either of two methods. The individual branch current values can be added, or Ohm’s law can be applied to the entire parallel circuit. Either method gives a total current of 12 A.

4. The total power taken by the entire parallel circuit is the product of the line voltage and the line current. (These values are in phase.)

The angle whose cosine is 1.00 is 0°. In other words, the line current is in phase with the line voltage.

6. The vector diagram for this parallel circuit is shown in Figure 8–2. The individual branch current values are placed on the voltage vector. Both the arithmetic sum and the vector sum of the branch currents equal the total line current.

**PARALLEL CIRCUIT WITH BRANCHES CONTAINING R AND XL**

Another type of parallel circuit has one branch containing a noninductive resistance load and a second branch containing pure inductance (Figure 8–3).

The two branch currents for this parallel circuit are 90 electrical degrees out of phase with each other. The current in the noninductive resistance branch is in phase with the line voltage. The current in the pure inductive branch lags the line voltage by 90°. Thus, the total line current can be found using vector methods.

**PROBLEM 2**

**Statement of the Problem**

For the parallel circuit shown in Figure 8–3, determine the following:

1. The current taken by each of the two branches

2. The vector diagram for the parallel circuit

3. The total current

4. The power taken by the parallel circuit

5. The power factor for the parallel circuit

6. The phase angle

7. The combined impedance

**Solution**

1. The voltage across each branch of the circuit is 120 V. The current in each branch of this circuit is as follows:

2. The current and voltage vector diagram for an RL parallel circuit is shown in Figure 8–4. The vector I is in the negative Y direction. Recall that in series circuits, the vector V was in the positive Y direction. The change in direction does not involve new concepts, but occurs because the reference vector has changed from I to V. The line current is the vector sum of the branch currents and lags behind the line voltage by an angle of 21.5°. (The calculation of this angle is given in steps 5 and 6.)

3. The total line current is really the hypotenuse of the right triangle in Figure 8–4, where

4. The power taken by the parallel circuit is used in the first branch containing resistance. No power is taken by the pure inductive load in the second branch. The power for the parallel circuit is

5. The power factor of any ac circuit is the cosine of the angle *fJ*. This value equals the ratio of the power to the input volt-amperes. The power factor is also the ratio of

the in-phase current to the total line current. For the circuit in Figure 8–4, the power factor is

6. The power factor angle (*fJ*) is 21.5° lagging.

7. The combined impedance of a parallel circuit is found using Ohm’s law. The impedance formula for a series circuit cannot be used because each parallel branch connected across the line wires tends to reduce the total parallel circuit impedance:

**Impedance Triangle for Parallel Circuits**

In series circuits, the voltage triangle is divided and multiplied by current to obtain the impedance and power triangles, respectively. In parallel circuits, the current triangle is divided by the reference vector V to obtain the impedance triangle (Figure 8–5).

The impedance triangle shown in Figure 8–5 for a parallel circuit does not resemble the series circuit impedance triangle. The impedance formula for a parallel circuit is

Another formula that can be used to determine the impedance of resistance and inductive reactance connected in parallel is

Assume that a parallel circuit contains a 15-n resistor connected in parallel with an inductor that has an inductive reactance of 20 n. To determine circuit impedance, substitute the values in the following formula: