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Unformatted text preview: capacitor after switching. Questions : 1. Q1 = C1V1; Q2 = C2V2; V0C1= Qtotal; Qtotal = Q1 + Q2; V0C1=C1V1 + C2V2 = CeqV; V1 = V2 = V; V0C1 = V(C1 + C2); V = V0C1/(C1 + C2) 2. Voltage decays because the capacitor is discharging. 3. 4. V1 = V0/C1 +C2 5. No we had faulty equipment. 6. V1 and V2 should equal V0 in series but should be equal in parallel Results: Known Capacitor Value (Parallel) (F) Voltage Across C1 (V) 1 0.2 0.04 2 0.36 0.04 3 0.48 0.04 4 0.56 0.04 5 0.64 0.04 6 0.72 0.04 Battery Voltage = 1.4 0.1 V Voltage across C1 in series: 0.64 V 0.04 Voltage across C2 in series: 0.64 V 0.04 Conclusion: Due to unfortunate circumstances and Stony Brooks need to fund landscaping instead of working equipment in laboratories, our experiment failed to give us the unknown capacitance. The equipment was most likely faulty due to the fact our instructor double checked our wiring and the oscilloscope did not read anything as expected....
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- Fall '08