Question

5) The energy stored in the capacitor in this circuit was zero at $t = 0$. Then, 2[s] later, the switch closed. Find $i_x(4[s])$.\newline$i_s = 50[mA]$\newline$R_1 = 10[\Omega]$\newline$t = 2[s]$\newline$C_x = 0.16[F]$\newline$R_2 = 43[\Omega]$\newline$i_x(t)$

Name: 5) The energy stored in the capacitor in this circuit was zero at t = 0. Then, 2[s] later, the switch closed. Find ix(4[s]).is = 50[mA]R1 = 10[Ω]t = 2[s]Cx = 0.16[F]R2 = 43[Ω]ix(t)
Uploaded: 2023-05-08T19:18:13-08:00
Duration: 5 min 35 s
Channel: Madhur L
Description: 5) The energy stored in the capacitor in this circuit was zero at t = 0. Then, 2[s] later, the switch closed. Find ix(4[s]).is = 50[mA]R1 = 10[Ω]t = 2[s]Cx = 0.16[F]R2 = 43[Ω]ix(t)

          5) The energy stored in the capacitor in this circuit was zero at $t = 0$. Then, 2[s] later, the switch closed. Find $i_x(4[s])$.\newline$i_s = 50[mA]$\newline$R_1 = 10[\Omega]$\newline$t = 2[s]$\newline$C_x = 0.16[F]$\newline$R_2 = 43[\Omega]$\newline$i_x(t)$

5) The energy stored in the capacitor in this circuit was zero at t = 0. Then, 2[s] later, the switch closed. Find ix(4[s]).is = 50[mA]R1 = 10[Ω]t = 2[s]Cx = 0.16[F]R2 = 43[Ω]ix(t)

Added by Vicenta C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

05/08/2023

Step 1

Since the energy stored in the capacitor at t=0 is zero, we can determine the initial voltage across the capacitor using the formula for energy stored in a capacitor: E = 0.5*C*V^2. Since E=0, V=0 at t=0. Show more…

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The energy stored in the capacitor in this circuit was zero at t=0. Then, 2[s] later, the switch closed. Find i_(x)(4[s]). please show all steps and expanations. Please DO NOT use laplace transformation. 5) The energy stored in the capacitor in this circuit was zero at t = 0. Then, 2[s] later, the switch closed.Find ix(4[s]) R =10[ =2[s] R,= 43[9] 50[mA] 0.16[F] x(t)