Example Parallel AC Circuit Problems
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Example Parallel AC Circuit Problems

Course Number: ECT 1225, Spring 2013

College/University: DeVry Chicago

Word Count: 646

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ECT125 Parallel AC Circuit Analysis The following are example problems calculated for parallel AC circuits. They include parallel RL, LC, and RLC circuits. The equivalent series circuit will be shown for all three configurations. An RC parallel combination circuit is not shown, but using the process for the RL circuit, you should be able to calculate an example of your own. The procedures shown are for calculating...

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AC ECT125 Parallel Circuit Analysis The following are example problems calculated for parallel AC circuits. They include parallel RL, LC, and RLC circuits. The equivalent series circuit will be shown for all three configurations. An RC parallel combination circuit is not shown, but using the process for the RL circuit, you should be able to calculate an example of your own. The procedures shown are for calculating ZT. You should be able to calculate total current, individual component currents, component powers, and the apparent power(PA) for the circuit. Practice putting all of your answers in polar and rectangular form and use the branch currents to verify Kirchoffs current law for total current in the circuit. Make sure you include your angles with the current values you calculate and put proper units on circuit parameters. Enjoy! Example 1: Find ZT for the circuit shown. (RL analysis) V1 8 Vrms 903 Hz 0 L1 ZT => R1 470mH 2k Figure 1. Parallel RL Circuit Solve for ZT by using product over sum: The denominator part of the formula is where you can use rectangular to polar conversion on your calculator. ZT = (2.67K)(2K) 90__________ = 5.34x106 (900) / 3.336K (53.160) ZT = 1.6K (90 53.16) = 1.6K 36.840 polar form Or solve by the reciprocal method: ZT = Converting the denominator to polar form we get: ZT = 1 / 6.247x10-4 -36.840 = 1.6K 36.840 Notice that 1.6K36.840 in polar form is equivalent to 1.28K + 959.3j in rectangular form. This says that the original parallel circuit can be transformed into a series circuit with a resistor of 1.28K and an inductive reactance of 959.3. The equivalent circuit would look as follows: R2 1.28k V2 L2 169mH 8 Vrms 903 Hz 0 Figure 2. E quivalent Series Circuit for Fig. 1 169mH is the equivalent inductance to get an XL = 959.3. This was found by: L = XL / (2f) = 959.3 / (2903Hz) = 169mH The same approach would be taken for a parallel RC circuit. Just remember that Xcj is negative or Xcj because Xc is represented 2: as. Example Find ZT for the circuit shown. (LC analysis) V3 20 Vrms 1kHz 0 L3 470mH C1 .1F ZT => Figure 3. Parallel LC Circuit Xc = 1 / (2(1KHz)(.1uF)) = 1.59 K XL = 2(1KHz)(470mH) = 2.953 K ZT = Zp = XL // Xc ZT = Notice that the denominator consists of 2 polar values that can be written in rectangular form. Therefore: ZT = As you can see, this is a capacitor. The circuit essentially looks like a capacitor where: V4 C2 .0462F 20 Vrms 1kHz 0 Figure 4. Equivalent Circuit for Fig. 3 So Xc = 3.445K with an equivalent C = .0462 uF. This is the equivalent circuit. Note that in the original circuit Xc was smaller than XL and the equivalent circuit reduced down to capacitive circuit. If Xc is larger than XL then the circuit would reduce down to an inductive circuit. This is an important concept to remember. Now lets look at an example of a parallel RLC circuit. Similar values are used as before to simplify the process. Example 3: Find ZT for the circuit shown. (RLC analysis) V3 20 Vrms 1kHz 0 L3 470mH C1 .1F R3 ZT => 5k Figure 5. Parallel RLC Circuit We use the same L and C as before and put a resistor (R) of 5K in parallel with the circuit. We already know that L // C is an equivalent capacitance where Xc = 3.445K. Now the circuit reduces down to: V4 C2 .0462F 20 Vrms 1kHz 0 Figure 6. Equivalent Circuit for Fig. 5 R4 5k Now solve the problem as an RC circuit using product over sum or the reciprocal method. This shows that three components in parallel can be broken down into a two part problem consisting of the resultant L //C taken in parallel with R. ZT = RXc / (R + Xc) = ZT = 2,837 -54.430 Polar form In rectangular form ZT = 1650 2307.6j This results in an equivalent series circuit as shown: R5 1650 V5 20 Vrms 1kHz 0 C3 .069F ZT => Figure 7. Equivalent Series Circuit for Fig. 6 The total current, component currents, and powers for the original components can now be calculated using Ohms Law and the power formulas.

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