Chapter 25b - Last time we developed the idea that...

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eq q eq C eq q + eq q Last time we developed the idea that combinations of capacitors can be reduced to an equivalent single capacitor with capacitance C eq and charge (separation) q eq . For capacitors in parallel: The rules we found were that: + 1 C 2 C 2 q 1 q V V V The voltage across each cap is the same , V 1 = V 2 =…= V . eq 1 2 q q q ... =++ & eq 1 2 C C C ... + V 1 C q, 2 C 1 V 2 V For capacitors in series: The charge across each cap is the same , . 12 V V V ... = ++ & eq 1 2 11 1 ... CC C = q + q e q q q ... q = ==
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All caps C =10 μ F All V = 10 V Find q 2 and V 2 for each circuit. Examples 1 2 3
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C 1 = C 2 =10 μ F V = 10 V Find q 2 and V 2 . C 1 C 2 0 V 10 V + + C eq = C 1 |C 2 0 V 10 V + “C 1 series C 2 2 12 eq 1 2 1 C C 100 F CC | C 5 F 11 2 0 F μ == = = = μ + Now find , eq eq q C V (5 F)(10V) 50 C = = μ Example 1 eq q
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C 1 = 10 μ F 0 V 10 V + C 2 = 10 μ F eq C5 F eq q5 0 C 0 V 10 V + Caps in series each have the same q i as q eq . 1 0 C = μ 2 0 C = μ Then, Looking for q 2 and V 2 , so must now work backwards 2 2 2 0 C V5 V C1 0 F μ == = μ Found: Note also that 1 1 1 0 C V 0 F μ = μ So, 12 V V 5V 5V 10V + =+= As it should for series caps. values are the answers.
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Time saving rules of thumb for 2 equal caps in series: C C 0 V V + eq C C 2 = V 2 V 2 The equivalent capacitance is half the capacitance of the equal capacitors. The total voltage drop splits evenly between the two caps.
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C 1 C 2 10 V + 0 V C 1 |C 2 C 3 10 V + 0 V C 3 0 V 10 V eq 1 2 3 C( C | C ) | | C = “parallel” C 1 = C 2 =10 μ F V = 10 V Find q 2 and V 2 . Example 2 “series”
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0 V 10 V eq 1 2 3 C (C |C ) ||C (10 F |10 F) ||10 F == μ μ μ eq C( 5 F ) | | 1 0 F = μμ eq C1 5 F = μ ( || just add) Calculate q eq , eq eq qC V ( 1 5 F ) ( 1 0 V ) = eq q 150 C = μ Calculate C eq ,
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0 V 10 V eq C1 5 F eq q 150 C C 1 |C 2 = 5 μ F 10 V + 0 V q C1|C2 = (5 μ F)(10V) q C1|C2 = (C 1 |C 2 )(V) q C1|C2 = 50 μ C q 1 = q 2 = 50 μ C C 3 = 10 μ F Now work backwards, Since series caps have the same charge.
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2 2 2 q5 0 C V5 V C1 0 F μ == = μ C 1 |C 2 = 5 μ F 10 V + 0 V C 3 C 1 C 2 10 V + 0 V C 3 expanding the left arm C 1 = 10 μ F 0 V 10 V + C 2 = 10 μ F 1 0 C = μ 2 0 C = μ left arm only As we had previously for 10 V across two 10 μ F caps in series.
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10 V + 0 V C 1 = 10 μ F C 2 = 10 μ F 1 q5 0 C 2 0 C C 3 = 10 μ F 33 3 q C V (10 F)(10V) 100 C = = μ For completeness we also find q 3 . Note that the charge for both arms is, 13 23 q q q q 50 C 100 C + =+= μ + μ eq 150 C q = μ= As we found above using C eq and V . 5V 5V 10V
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10 V 0 V 10 V 0 V C 2 C 3 C 1 C 4 C 4 C 2 |C 3 C 1 Concentrate on this side since other side will not affect q 2 and V 2 .
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Chapter 25b - Last time we developed the idea that...

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