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Unformatted text preview: Of the four circuits in Part D (4), circuits a and c can be solved using the equation presented in class, because the circuits can be reduced by finding the Thevenin equivalent. Vc(t) = K1 + K2 x e^(t/RC) Vc( 4 ) = K1 = Vf K1 = Vf Vc(0) = Vf + K2 = Vi K2 = ViVf Measured Values R1 = 98.695 kOhm R2 = 98.944 kOhm R3 = 9.914 kOhm C1 = 392.1 pF C2 = 369.1 pF Circuit A Vc(t) = Vf + (ViVf) x e^{t/[(R1+R3) x C2]} = Vf + (ViVf) x e^(t/4.01 x 10^5 sec) Circuit B Vc(t) = Vf + (ViVf) x e^{t/[(R1+R3R2) x C2]} = Vf + (ViVf) x e^(t/1.91 x 10^5 sec) 4) See attached page for table Circuit C Vout = [R2/(R2+R1+R3)]Vin = [98.944/(98.944+98.695+9.914)] x 5V = 2.38V Circuit D C2 Vout = [R2/(R2+R1+R3)]Vin = [98.944/(98.944+98.695+9.914)] x 5V = 2.38V C1 Vout = [(R1(R2+R1+R3]Vin = [98.695/(98.944+98.695+9.914)] x 5V = 2.38V *References results from lab notebook on last two pages of report...
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 Spring '09
 Zhang
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