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Unformatted text preview: After switch is thrown to B: V1 = V23 (parallel branches)
q2 and q3 in series: q2 = q3 = q23 (C23 = 2μF)
Charge conservation: q10 = q1 + q23
48 = C1V1 + C23V1 (V1 = V23)
Find V1: V1 = 48 / (C1 + C23) = 8 V
Find q1: q1 = C1V1 = 32μC
12V
q23 = q2 = q3 = 48 – 32 = 16μC
V2 = q2 / C2 = 2.67 V
V3 = q3 / C3 = 5.33 V PHY2049: Chapter 25 A B
6μF 4μF 3μF 24 Another Example
Each capacitor has capacitance 10μF. Find the total
capacitance
Do it in stages
2&3
Add 4
Add 5
Add 1 ⇒
⇒
⇒
⇒ 5 μF
15 μF
6 μF
16 μF 2
1 4 3 5 PHY2049: Chapter 25 25 Find Charges on All Capacitors
Each capacitor has capacitance 10μF. V = 10 volts
q1 = 10 x 10 = 100μC C2345 = 6μF
q2345 = q234 = q5 (series)
q2345 = q234 = q5 = 10 x 6 = 60μC
V5 = q5 / C5 = 6 Find q4, V4
V234 = V4 = 10 – 6 = 4
q4 = C4 x 4 = 40μC 2
1 4 3 5 Find q2, q3, V2, V3 (C23 = 5μF)
q2 = q3 = q23 = C23 x 4 = 5 x 4 = 20μC
V2 = 20 / C2 = 2
V3 = 20 / C3 = 2
PHY2049: Chapter 25 26...
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This note was uploaded on 07/31/2011 for the course PHY 2049 taught by Professor Any during the Summer '08 term at University of Florida.
 Summer '08
 Any
 Physics, Capacitance, Charge, Electric Fields

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