1 1 AM && (100000) x ( 470 ×10*) = 47, time что и 100 y Changing time (s) (valt) dischay t time (3) " 11-75 2-5 2.3 5.5

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1 1 AM && (100000) x ( 470 ×10*) = 47, time что и 100 y Changing time (s) (valt) dischay t time (3) " 11-75 2-5 2.3 5.5 #1 23.5 4.1 3.7 5 35.25 5.3 2.6 "r 7.5 47 " 6.2 1.7 10 " 70-5 7.1 0.8 15 94 7.6 0.3 20 117-5 7.8 0.0 25 141 7.9 O 36 164.5 7.9 0 35 40 188 7.9 211-5 45 235 258.5 8 282 470 T D. 25 0.50 0.75 1.0 1.50 2 2.5 3 3-5 4 4.5 S 5.5 6 do do 8 Oo 0 oo 0 0 8 0 : If 3 (+ : " " " fi ( # 50 55 60 (100.000) x (100×10) 105 differing 6.2 4.5 3.2 2.2 I 0.4 d. 2 Ə O d 0 0 105 Marting 1.8 39 5 5.9 ㅋ 7.5 7-7 100 7.9 7.9 8 8 8 8 8 0 C
V. GUIDE QUESTIONS FOR ANALYSIS AND DISCUSSION 1. Compare the time it takes to fully charge the capacitor in trial I and trial 2. 2. The voltage across the capacitor, Vc, at any instant time during the charging period is given as: V. = V₁ (1-e¹/RC) where V, is the voltage supply R is the resistance in Ohm C is the capacitance in Farad t is the time elapsed since the application of the supply a. Using the equation above, what is the voltage across the capacitor in trial I after 10 seconds? b. From the graph, approximate V, after 10 seconds in trial 1 and compare it to the computed value. 3. Given the circuit below t=0 R = 100ΚΩ www Vc = 10v 22uF The discharging voltage across the capacitor, Ve, at any instant time during the discharging period is given as: V₁ = V₁ e-U/RC where V₁ is the initial voltage R is the resistance in Ohm C is the capacitance in Farad Activity 05 Voltage Variations in Capacitors Page 3 t is the time elapsed since the application of the supply a. Compute for the time constant b. What is the voltage across the capacitor after 5 seconds?