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Consider three capacitors having capacitances of 3.30 µF, 8.65 µF, and 8.40 µF. Find their equivalent capacitance if they are connected as follows. (a) in parallel Ceg = (b) in series Ceg =

Consider three capacitors having capacitances of 3.30 µF, 8.65 µF, and 8.40 µF. Find their equivalent capacitance if they are connected as follows. (a) in parallel Ceg = (b) in series Ceg =

Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
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Example 26.4 Equivalent Capacitance Problem Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 26.11a. All capacitors are in microfarads. 4.0 4.0 13.0 Cea Strategy Using Equations 26.8 and 26.10, we reduce the combination 8,0 8.0 step by step as indicated in the figure. (a) (b) (c) (d) Figure 26.11 To find the equivalent combination of the capacitors in (a), the various combinations are reduced in steps as indicated in (b), (c), and (d), using the series and parallel rules described in the text. All capacitance values are in microfarads. Solution The C = 2.0 µF and 3.0 µF capacitors are in parallel and combine according to Ceg = C1 + C2. Their equivalent capacitance is x1 = 5.0 µF. Likewise, the C2 = 4.0 µF and 6.0 µF capacitors are also in parallel and have an equivalent capacitance of y1 = 10.0 µF. The upper branch in Figure 26.11b now consists of a 4.0 µF capacitor and a 5.0 µF in series, which combine to give x2 according to the following equation. 1 1 1 1 1 C1 C2 4.0µF 5.0µF x2 X2 = Likewise, the lower branch in Figure 26.11b consists of an 8.0 µF capacitor and a 10.0 µF in series, which gives what equivalent capacitance? Y2 = µF Finally, the two capacitors in Figure 26.11c are in parallel and have what equivalent capacitance, which is also the equivalent capacitance of the circuit? Ceg = You can practice reducing a combination of capacitors to a single equivalent capacitor with this Interactive Example. Exercise 26.4 Hints: Getting Started | I'm Stuck Consider three capacitors having capacitances of 3.30 µF, 8.65 µF, and 8.40 µF. Find their equivalent capacitance if they are connected as follows. (a) in parallel Ceg = µF (b) in series Ceg = µF
Transcribed Image Text:Example 26.4 Equivalent Capacitance Problem Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 26.11a. All capacitors are in microfarads. 4.0 4.0 13.0 Cea Strategy Using Equations 26.8 and 26.10, we reduce the combination 8,0 8.0 step by step as indicated in the figure. (a) (b) (c) (d) Figure 26.11 To find the equivalent combination of the capacitors in (a), the various combinations are reduced in steps as indicated in (b), (c), and (d), using the series and parallel rules described in the text. All capacitance values are in microfarads. Solution The C = 2.0 µF and 3.0 µF capacitors are in parallel and combine according to Ceg = C1 + C2. Their equivalent capacitance is x1 = 5.0 µF. Likewise, the C2 = 4.0 µF and 6.0 µF capacitors are also in parallel and have an equivalent capacitance of y1 = 10.0 µF. The upper branch in Figure 26.11b now consists of a 4.0 µF capacitor and a 5.0 µF in series, which combine to give x2 according to the following equation. 1 1 1 1 1 C1 C2 4.0µF 5.0µF x2 X2 = Likewise, the lower branch in Figure 26.11b consists of an 8.0 µF capacitor and a 10.0 µF in series, which gives what equivalent capacitance? Y2 = µF Finally, the two capacitors in Figure 26.11c are in parallel and have what equivalent capacitance, which is also the equivalent capacitance of the circuit? Ceg = You can practice reducing a combination of capacitors to a single equivalent capacitor with this Interactive Example. Exercise 26.4 Hints: Getting Started | I'm Stuck Consider three capacitors having capacitances of 3.30 µF, 8.65 µF, and 8.40 µF. Find their equivalent capacitance if they are connected as follows. (a) in parallel Ceg = µF (b) in series Ceg = µF
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