where qo is the maximum charge on the plates and is the capacitive time constant (>=...
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where qo is the maximum charge on the plates and is the capacitive time constant (>= RC, where R is resistance and C is capacitance). NOTE: The stated value of a capacitor may vary by as much as 20% from the actual value. Taking the extreme limits, notice that when t = 0, q=0 which means there is not any charge on the plates initially. Also notice that when t goes to infinity, q goes to qo which means it takes an infinite amount of time to completely charge the capacitor. The time it takes to charge the capacitor to half full is called the half-life and is related to the capacitive time constant in the following way: t = RC In 2 In this activity the charge on the capacitor will be measured indirectly by measuring the voltage across the capacitor since these two values are proportional to each other: q=CV 5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 9.70 -0.5 Voltage (V) A of 9.80 9.90 1 1 I 1 1 1 Fit 0.2600 Voltage Graph 9.8800, 3.052E-3) 10.00 Data 10.10 2.017 10.20 Time(s) 10.30 X Voltage, ChA Run #8 10.40 10.50 10.60 Data Resistance of the resister (R) = 1000 Ohms Given value of the Capacitance of the Capacitor(C) = 375 F Time taken to reach half of the maximum Voltage (from graph), t/2 = 0.26 s Questions 1. Calculate the capacitance of the capacitor: /2 RC In 2= RC (0.693) Capacitance 2. Determine the percent difference between the calculated capacitance and the value on the capacitor (330 uF): 3. % diff= LUF calculated-330F 330F The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how long does it take for the capacitor to charge to 75% of its maximum? V V. - v (1-2 The) RC x 100 % where qo is the maximum charge on the plates and is the capacitive time constant (>= RC, where R is resistance and C is capacitance). NOTE: The stated value of a capacitor may vary by as much as 20% from the actual value. Taking the extreme limits, notice that when t = 0, q=0 which means there is not any charge on the plates initially. Also notice that when t goes to infinity, q goes to qo which means it takes an infinite amount of time to completely charge the capacitor. The time it takes to charge the capacitor to half full is called the half-life and is related to the capacitive time constant in the following way: t = RC In 2 In this activity the charge on the capacitor will be measured indirectly by measuring the voltage across the capacitor since these two values are proportional to each other: q=CV 5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 9.70 -0.5 Voltage (V) A of 9.80 9.90 1 1 I 1 1 1 Fit 0.2600 Voltage Graph 9.8800, 3.052E-3) 10.00 Data 10.10 2.017 10.20 Time(s) 10.30 X Voltage, ChA Run #8 10.40 10.50 10.60 Data Resistance of the resister (R) = 1000 Ohms Given value of the Capacitance of the Capacitor(C) = 375 F Time taken to reach half of the maximum Voltage (from graph), t/2 = 0.26 s Questions 1. Calculate the capacitance of the capacitor: /2 RC In 2= RC (0.693) Capacitance 2. Determine the percent difference between the calculated capacitance and the value on the capacitor (330 uF): 3. % diff= LUF calculated-330F 330F The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how long does it take for the capacitor to charge to 75% of its maximum? V V. - v (1-2 The) RC x 100 %
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