Question

Review | Constants A 0.185 H inductor is connected in series with a 89 2 resistor and an ac source. The voltage across the inductor is ui= -(11.0V) sin(490rad/s)t]. Part A Derive an expression for the voltage ur across the resistor. Express your answer in terms of the variables L, R, VL (amplitude of the voltage across the inductor), w, and t. O ALQ * R 0 a ? UR = vr= Submit Previous Answers Request Answer Part B What is vr at 2.10 ms? Express your answer with the appropriate units. ? CH HÅR vrt Value O Units Submit Request Answer Provide Feedback Next >

Answer 1

Given L= 00185H , R = 892 * = -11 sin (490t) = +11 Sin (t + ugot) as sin (+8)= -Sino V = 11 sin (ygot+7) comparing with V2 = U, sin helt +Ø) - V=110, W = 490 radls 0 = π rad. (A) In resistor , voltage lags II than inductor so VR = VR sin(uqot + | -) - Ve Sin(490t+?) Now Up = IX2 = I WL, VR = IR =) VL 89 - = 100185 & VR = k 1.0185 - =) VR = 1 1 1 1 1 85 = 10.8 =) VR = VR sin (wt+II) where VR in terms of variable - W2 - ve- RVL ve = L = 490X0.185 بالا v + Raya Sinfurt + 9 t = 2 Lo 73 = 0:00 Qlo ý , VR = % Sout-T) - .8 SA (1 fº X 0-00 đ = 15.57 v