What is the voltage for each capacitor ifplates of opposite sign are connected?
ID: 1672084 • Letter: W
Question
What is the voltage for each capacitor ifplates of opposite sign are connected?What is the charge on each capacitor if plates of opposite sign areconnected?
A 2.80 mu F capacitor is charged to 460 V and a3.75 mu F capacitor is charged to 525 V. These capacitors are then disconnected from their batteries, andthe positive plates are now connected to each other and thenegative plates are connected to each other. What will be thepotential difference across each capacitor? What will be the charge on each capacitor? What is the voltage for each capacitor ifplates of opposite sign are connected? What is the charge on each capacitor if plates of opposite sign areconnected?
Explanation / Answer
Given C1 = 2.80 * 10-6 F C2 = 3.75 * 10-6 F V1 = 460 V V2 = 525 V a. When capacitors are connected such that + ve plate connects to the +ve plate, equilibrium potential difference V++ = (C1 * V1 + C2 * V2) / (C1 + C2) = (2.80 * 10-6 * 460 + 3.75 * 10-6 * 525) / (2.80 * 10-6 + 3.75 * 10-6) = 497.21 V b. Charge on C1, Q1 = C1 * V++ = 2.80 * 10-6 * 497.21 = 1.39 * 10-3 C = 1.39 mC Charge on C2 Q2 = C2 * V++ = 3.75 * 10-6 * 497.21 = 1.86 mC = 3.75 * 10-6 * 497.21 = 1.86 mC c. When capacitors are connected with opposite polarity, equilibrium potential difference is V+- = (C1 * V1 ~ C2 * V2) / (C1 + C2) = (2.80 * 10-6 * 460 ~ 3.75 * 10-6 * 525) / (2.80 * 10-6 + 3.75 * 10-6) = 103.93 V = 2.80 * 10-6 * 103.93 = 2.91 * 10-4 C = 0.291 mC Charge on C2 Q2' = C2 * V+- = 3.75 * 10-6 * 103.93 = 3.90 * 10-4 C = 0.390 mC = 3.75 * 10-6 * 103.93 = 3.90 * 10-4 C = 0.390 mCRelated Questions
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