Use the graph to determine the time to 'half-max'. t1/2 = S Calculate the capacitance (C)...

  

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Use the graph to determine the time to 'half-max'. t1/2 = S Calculate the capacitance (C) of the capacitor. C = μF Determine the percent error between the calculated capacitance and the value on the capacitor (330 μF). calculated 330 μF 330 μF % error = * 100% = % What is the maximum charge for the capacitor in this experiment (approximately)? Charge equals capacitance divided by voltage, or 0.000330 F/4.00 V, or 0.00132 C or 1320 μC. Charge equals capacitance multiplied by voltage, or 330 F x 4.00 V, or 1.32 C. Charge equals capacitance multiplied by the inverse of the voltage, or 0.000330 F x 1/4.00 V, or 0.0000825 C or 82.5 µC. Charge equals capacitance multiplied by voltage, or 0.000330 F x 4.00 V, or 0.00132 C or 1320 μC. 80 OF of 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 9.70 -0.5 Voltage(V) A 9.80 d Fit F.90 0.2600 (9.8800, 3.052E-3) Voltage Graph 10.00 ΑΣ· & Data 2.017 10.10 10.20 Time(s) 10.30 X Voltage, ChA Run #8 10.40 10.50 10.60 Use the graph to determine the time to 'half-max'. t1/2 = S Calculate the capacitance (C) of the capacitor. C = μF Determine the percent error between the calculated capacitance and the value on the capacitor (330 μF). calculated 330 μF 330 μF % error = * 100% = % What is the maximum charge for the capacitor in this experiment (approximately)? Charge equals capacitance divided by voltage, or 0.000330 F/4.00 V, or 0.00132 C or 1320 μC. Charge equals capacitance multiplied by voltage, or 330 F x 4.00 V, or 1.32 C. Charge equals capacitance multiplied by the inverse of the voltage, or 0.000330 F x 1/4.00 V, or 0.0000825 C or 82.5 µC. Charge equals capacitance multiplied by voltage, or 0.000330 F x 4.00 V, or 0.00132 C or 1320 μC. 80 OF of 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 9.70 -0.5 Voltage(V) A 9.80 d Fit F.90 0.2600 (9.8800, 3.052E-3) Voltage Graph 10.00 ΑΣ· & Data 2.017 10.10 10.20 Time(s) 10.30 X Voltage, ChA Run #8 10.40 10.50 10.60