Three capacitors of capacitance C1 -2.90 µF and C2 = 2.20 µF and C3 = 4.25 µF ar

Question

Three capacitors of capacitance C1 -2.90 µF and C2 = 2.20 µF and C3 = 4.25 µF are connected in parallel, as illustrated in the Active Figure after adjusting to display three capacitors. A potential difference of ?Vb = 98.0 V is maintained by a battery. (A) Find the charge Q stored by the assembly of capacitors. (B) Find the equivalent capacitance of the assembly of capacitors. (C) A set of switches is suddenly thrown to disconnect the battery and interchange the positive and negative sides of C2. Find the new voltage across the capacitors.

Explanation / Answer

b) total capacitance = 2.9 + 2.2 + 4.25 = 9.35 F

a) Q = CV = 9.35 * 98 = 916.3 C

c) when C2 is reversed, it will lose its charges to C1 and C3.

Q2 = 2.2 * 98 = 215.6 C

Q1+3 = 916.3 - 2(215.6) = 485.1 C

Q1 = 2.9/(2.9 + 4.25) * 485.1 = 196.75 C

Q3 = 4.25/(2.9 + 4.25) * 485.1 = 288.35 C

V1 = 196.75 / 2.9 = 67.84 V

V2 = 0

V3 = 288.35 / 4.25 = 67.85 V

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