Question

Please show any equation manipulation.

Problem 2 HE In the given RC circuit, a capacitor is connected to a resistor in series and is getting charged after closing the Switch. The time constant of the circuit is 10 (S). R = 109 (0) C =? (F) A- Calculate the capacitance of the capacitor. B- How much time does it take for the capacitor to become fully (about 99%) charged? C- If we close the Switch and the capacitor starts getting charged, what is the charge at the 5th second of Switch being closed? D- How much time does it take after closing the Switch for the current in the circuit become 0.5X10* (A)? Switch E- What is voltage across the capacitor at the 5th second V = 10 (V) of Switch being closed? F- What is the energy stored in capacitor at the 5th second of Switch being closed? { -

Answer 1

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ma Vb = VR TV IR TV Fc after solving by izda on capacitor , Vcx) = V CI carrent Voltage charge on capaciter = qui a coleta) -t at = 1(C) = Vb t e RC - Rc A- time constant =los RC = 10 cz lo =108 न 109 aman -t Br - CVb at 99% q=0.99 CVb 0.99 (+6 = CV Cl- eTo To), t=46.0515 c- at t=55 a= CVbcl-etto) = 10% loci-esto q 3.93x108 c d=55 D- time for 10 - 0.5*108 0.5x108 t= 6.935 107 (1-00.5) = lo t - 109 xe To

tr 108 1) E = energy stored stcu² E- Vollege at 5s charge at ss Capacitance v 3:93x108 d=5s = 3.93 V ₂C Ves t=5s t=5s = 0.5x108x x 3.932 Exass - 7.72*10% ] 그 다 2