Question

An RC-circuit (Fig.1) contains components R and that need to be determined. You have a digital multi-meter (DMM) that can measure resistance (R), Voltage (V).etc., but not capacitance (C). To measure is not possible, 50 You can determine c from the time-courtone G-RC of the circuit. For this, you design av experiment: A known voltage V (say, Vo=97 from a qv batery) 25 applied to the circuit, then you mearure the voltage across the capacitor, say, you necard valt) every 5 seconds, and obtain the fallawang table time t 5] 5 10 15 20 25 30 35 40 45 50 Voltage Ccct) [v1 o 1.82 3.30 14.40 5.4 5.9 6.75 7.1 17.5 7.85 9.1 (TC.) mecoured data (natsy) - R=lokuu 4.3.- Sct Vocco [] dato V- 9V CE? + (4) sw ted 19.0: RC Circuit . Switch O is clued at t= %. Capacitor 0 5 10 15 20 25 30 -.. is initially-discharged, 4(0)-. Fy & Measurement of (t) at timer t= 5,10,15,20, etc. [+], and You know that the capacitor responde datted-Live) will charge according to (t)= Vot-et/s) where GERCfrom which the copacitance of the capacitor is Cabo Jd you have measured R to be R=47k15 , then by knowing G Yousiudc But, how do we determine ? - Simple!: From @when to the capacitor voltage is 05 (2)= Vo (1-2)=0.632 V 9

Answer 1

v=qv v (2) = Voll-e ") = 0.632 Vo = 0.632 (91) = 5.7 a as we estimate from data the 20+25 - z 1 Because S.TV is between 5.4 at t=205 and [ 5.9 at t= 255 e ~22:55 T=Rc 22.55= (47kr) c -22.55 = 47X103 c=478.72 UF