1Bch24 - Chapter 24 There are two main types of capacitor...

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Unformatted text preview: Chapter 24 There are two main types of capacitor configurations: series and parallel. For each of these, here’s how the charges and voltages across each capacitor are related, and how to compute the equivalent capacitance. Capacitors in Series: Consider capacitors C 1 , C 2 , ......., C N arranged in series, i.e., strung together. Then each capacitor stores the same charge: Q 1 = Q 2 = Q 3 = ... = Q N = Q The total potential V across the combination is the sum of the voltage V j across each capacitor C j : V = N X j =1 V j = N X j =1 Q j C j = Q N X j =1 1 C j = Q C eff Therefore, the effective capacitance of the combination is obtained by adding the inverses of the individual capacitances: 1 C eff = N X j =1 1 C j Capacitors in Parallel: Consider capacitors C 1 , C 2 , ......., C N arranged in parallel. Then the voltage across each capacitor is the same: V 1 = V 2 = V 3 = ... = V N = V The total charge Q stored in the combination is the sum of the charge Q j stored in each capacitor C j : Q = N X j =1 Q j = N X j =1 C j V j = V N X j =1 C j = V C eff Therefore, the effective capacitance of the combination is obtained by adding the individual capacitances: C eff = N X j =1 C j 1 Problem 1: Capacitors C 1 , C 2 , C 3 , C 4 , and C 5 are arranged as shown in the diagram below. A voltage V ab is applied between terminals a and b. 1: Find the equivalent capacitance of the system. 2: Find the charge and voltage across each capacitor. & & ¡ & ¢ & £ & ¤ & ¡¢ £ ¡ ¢ Solution: In order to compute the equivalent capacitance, we successively identify series and parallel combinations, simplifying the circuit as we go along. We first note that capacitors C 3 and C 4 are arranged in series, so their equivalent capacitance C 34 is given by 1 C 34 = 1 C 3 + 1 C 4 C 34 = C 3 C 4 C 3 + C 4 We now note that the equivalent capacitor C 34 is in parallel with capacitor C 2 , allowing us to find the equivalent capacitance C 234 of capacitors C 2 , C 3 , and C 4 . C 234 = C 2 + C 34 = C 2 + C 3 C 4 C 3 + C 4 = C 2 ( C 3 + C 4 ) + C 3 C 4 C 3 + C 4 Finally, we note that capacitors C 1 , C 234 , and C 5 are arranged in series. This allows us to 2 find the equivalent capacitance C eq of the system: 1 C eq = 1 C 1 + 1 C 234 + 1 C 5 C eq = • 1 C 1 + 1 C 234 + 1 C 5 ‚- 1 = • 1 C 1 + C 3 + C 4 C 2 ( C 3 + C 4 ) + C 3 C 4 + 1 C 5 ‚- 1 We now consider part 2 . The overall charge stored in the system is Q = C eq V ab This is the charge stored in C 1 , C 234 , and C 5 , since these capacitors are in series. That is, Q 1 = Q 5 = Q 234 = Q = C eq V ab We can now compute the voltages across capacitors C 1 , C 5 , and C 234 : V 1 = Q 1 C 1 = C eq V ab C 1 V 5 = Q 5 C 5 = C eq V ab C 5 V 234 = Q 234 C 234 = C eq V ab C 234 V 234 , however, is the voltage across capacitors C 2 and C 34 , since these capacitors are arranged in parallel. That is, V 2 = V 34 = V 234 = C eq V ab C 234 We can now compute the charges across capacitors C 2 and C 34 : Q 2 = C 2 V 2 = C 2 C eq V...
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1Bch24 - Chapter 24 There are two main types of capacitor...

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