Tarefasb_provaiv

4) q(t) = Q_max * cos(yt + ß); \u03B5 = 0 \nq(0) = Q_max * cos(0) \nQ_max = q(0) = \u03A8 = cos^{-1} (\\frac{q(0)}{Q_max}) \n\=> \u03A8 = 0.75 rad\n5) q(t) = -\\u03C9_0 * Q_max * sin(\\u03C9_0t + ß) \n\\=> \\u03C9_0 = 1/\\sqrt{LC} \n\\=> q(t) = -4.79*10^{-10} * sin(54.1^{\\circ} + 0.75) \n\\=> I(t) = (I_{5.10^{-10}) = 7.1089 * 10^{-3} A \n\\=> q(63.7^{\\circ}) (Exercício L2) \nQ_max = 7.1089 * 10^{-3} A\n6) q(10) = 8.169 * 10^{-5} C\nq(0) = 8.169 * 0.6673E\nq(0) = 8.169 * 10^{-6} C\n(I(10) = 8.829 * 10^{-6} C) 1) \\Phi_B = N \\cdot B \\cdot L \\cdot \\frac{1}{N} => \\Phi_B = 5.8.81 = \\Phi_B = 0.120715Wb => \\Phi_B = 12.0.15mWb\n2) \\text{Substituted: } \\Phi_B = N \\cdot B \\cdot A \\cdots(\\text{area}) => N = N \\cdot B \\cdot A \\cdots\\} \\implies N\\cdot B= BA => \\Phi_B = N\\cdot \\mu_0 \\cdot I_T \nV = \\text{solenoids}\n3) I_{fus = \\Phi_B}{\\cdots(\\text{of solenoid})}\nfem = -L \\cdot \\frac{d}{dt}(N\\cdot B A) => \\Phi_B = \\int N\\cdot B A \\cdots(\\text{add})\nfem = N \\cdots(B \\cdot \\mu_0) + \\text{ of induction} 1) \\Phi_B = (\\frac{8 \\pi \\cdot \\mu_0 R^4}{2x^3})\n2) \\frac{\\partial B}{\\partial x} = -\\frac{\\partial}{\\partial t}[\\mu_0(B \cdot R^4 \\cdot \\frac{1}{2x^3})]\\cdots{\\int}[\\frac{\\partial I_{ind}}{\\partial t}]/dt\n3) \\Phi_B = \\int dB \\cdots = \\int I \cdots \\int \\cdots\ne = \\mu_0 \\cdot \\epsilon dt => I(t) = \\frac{3.6313.10^{-2}+0.34.10^{-6}}{2.213} Cuestionario 17(b) (Circuito RLCa)\n1) E_trans = Z_trans (E_trans = E)\n\n E_trans = ∑ I_trans (E_trans = E)\n\nE_trans = ε / √2·Z = ∑ I_trans • R + (Xl - Xc)\n\n2) ΔV₁ = R · I\n\nΔV₁ = 20 · 0.0340837\nI₁ = 0.053V/\n\n3) ΔV₃ = Xl · I\n\nΔV₃ = 1 / ωC => ΔV₃ = 0.0340837\n\n200 · 9 · 10⁻⁶\n\nΔV₃ = 18.935V/\n\n5) ΔVs = ΔV / √2 => ΔVs = 25 => ΔVs = 17.678V/\n\n6) I_rms =\n\n 25 => I_rms = 0.04974087A\n\nΔV₁ = R · I\n\nΔV₁ = 20 · 0.03474087 => ΔV₁ = 10.2 · 9.81/\n\n7) ΔV₂ = Xc · I\n\nΔV₂ = 1 / ωC => ΔV₂ = 0.0340837\n\nΔV₂ = 200·9·10⁻⁶ ΔV₃ = Xc · I\n\nΔV₃ = 1 / ωC , I => ΔV₃ = 0.04974087 => ΔV₃ = 55.627V/\n\n1000 · 9 · 10⁻⁶\n\nΔV₂ = A