# hw11_problem5_yourlogin - 0.57 3 0.788 0.34 4 0.336 0.15...

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Name(s) : Purdue Email Address(es): Section #: Assignment #: I/We have not used material obtained from any other unauthorized source, ei or unmodified. Neither have I/we provided access to my/our work to another The project I/we am/are submitting is my/our own original work. Problem Description: Input Section: Output Se Time (sec) Voltage (V) 0 10 1 6.1 2 3.7 3 2.2 4 1.4 5 0.8 6 0.5 7 0.3 8 0.2 9 0.1 10 0.07 11 0.03 Calculation Section: General Form: General Form: Linearized Data: Linearized Form: Time (sec) ln (Voltage) log(Voltage) 0 2.303 1 1 1.808 0.79 2 1.308

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Unformatted text preview: 0.57 3 0.788 0.34 4 0.336 0.15 5-0.223-0.1 6-0.693-0.3 7-1.204-0.52 8-1.609-0.7 9-2.303-1 10-2.659-1.15 slope-0.496-0.215 voltage = b e m*time 2 4 6 8 10 12 D Voltage (V) intercept 2.303 1.000 general form slope-0.496-0.22 general form intercept 10.000 10.000 ither modified r. ection: ln(voltage) = -0.496 time + 2.303 voltage = 10 e-0.496 time 2 4 6 8 10 12 2 4 6 8 2 f(x) = 10.440.6^x R = 1 Discharge of a Capacitor Over Time Time (sec)...
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## This note was uploaded on 05/02/2011 for the course ENGR 132 taught by Professor Kutson during the Spring '11 term at Purdue University-West Lafayette.

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hw11_problem5_yourlogin - 0.57 3 0.788 0.34 4 0.336 0.15...

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