Question

In the figure, R1 = 12.0 12, R2 = 9.00 2, R3 = 4.502, ε = 63.0 V, and C = 6.00 uF. The capacitor is initially uncharged. The switch is closed at t = 0. a) Immediately after the switch is closed: i. What is the current through each resistor? ii. What is the potential across each resistor? iii. What is the potential across the capacitor? (10 points) b) After the switch has been closed for a long time, what is the charge on the capacitor and the current through each resistor? (5 points)

Answer 1

Page 1 using Theren in Theary in AB part me to RAR, Let R = RahtR3 9412 50, 53 = Io e ked where 10 = Este . Rek 2X12 = 316 ~ 5.143.2 Ex= 9x63 = 27 volt R = 4.5+36 4201 9:6422 So, Io = 27 xha = 2.8 A So, Iz = 2.8 e-te - 2.8 e-learning * 15-230° ted] =6verter 2. se

Page 2 Using KV2 in Fig-1 ist loop we get E = I, R, & Izka => 63 =123, + 91 — ☺ Using KCL in Fig i at junction ... we 1. I = Iz t I3 ③ From @. and ③ we get :63 = 12(32+79) +972 + 2132 + 1253 7 I = 63 +12 33 21 Iz = 3.0-422 23 using 0 in @ we get Iz = 3 +4 +2:8 he he → 12 = 361.62 putting 6 in ③ we get I = 3* 1 / 2 13 tI3 *1 = But that I, = 3 + 3 Iz => 1 = 3+1.2 et ©

Page=3 o current through R, is 19 = 3+1.2 e- A??. R2 is I2 = 3-1.6 e- isso e ng is I3 = 2.8.e-493 e A. where @= 6x10-6 F 11 1 4 = 17.28x103) (1) V=IR, using this. Cohm's law) potential aerosso R; es v, = I, Ri ., = 36+14.4 e-hande volt ! 35 Rais /2 = 27-14.96- 1914 e vesti Kз іѕ v, - Сез N = 12.6 e-letne vort 7 Ciii) currient through capacitor is Is . For capacitor, azeve . Canael » con le Ig » Ne = & first کا

Page 4 * več ne ha det 3 5 > Ve = 135. [l-e-ho > Ve = 33.75 (1--94) less 1350 32 to e ©. If feme 'f's large then, eine ne tends ko pero ne lim eat o her defer all a to - here a = ls is so ..so, in this case, voltage on capacitor is Ve = 33.75 se; charge. q = CV = 33:75x6x10-6 77% = 2.025100% and From set of egen ② in page -3 we get current through .R, is I =3A. . R2 is I = 3A. R3 is Iz=0A.