Lesson_3.2_Homework_and_Solutions

Lesson_3.2_Homework_and_Solutions - Lesson 3.2 Homework...

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Lesson 3.2 Homework Solutions 1. (i) The lowest value is obtained when they are connected in series. 1 2 3 1 1 1 1 1 1 1 0.217 10 15 20 4.6 C C C C C F μ = = + + = + + = (ii) The highest value is when they are in parallel. C = C 1 + C 2 + C 3 = 10 + 15 + 20 = 45 μ F 2. (i) 2 μ F 3 μ F Q -Q Q -Q 9 V V 1 V 2 (ii) Total capacitance is: 1 2 1 2 2 3 6 1.2 2 3 5 C C C F C C × = = = = + + Q = CV = 1.2 μ F × 9.0 V = 10.8 μ C (iii) Since they are in series, each capacitor has a charge 10.8 μ C. If V 1 is the potential difference across C 1 and V 2 across C 2 , we have 1 1 2 2 10.8 2 5.4 3 10.8 3 .6 Q C V C F Q C V C F V V = = = = = =
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(iv) 2 μ C 3 μ C 9 V 9 V 9 V Q Since they are in parallel, the potential difference across each capacitor is the same, namely 9.0 V. If Q 1 is the charge on C 1 and Q 2 on C 2 , we have: Q 1 = C 1 V = 2 × 10 -6 μ F × 9.0 V = 18 × 10 -6 C = 18 μ C Q 2 = C 2 V = 3 × 10 -6 μ F × 9.0 V = 27 μ C 3. When the two capacitors are in parallel, 35 = C 1 + C 2 …(i) This gives C 2 = 35 – C 1 When they are in series, 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 ...( ) 4 1 1 1 35 35 4 35 (35 ) (35
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_3.2_Homework_and_Solutions - Lesson 3.2 Homework...

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