In a simple AC circuit shown on the right, C = 0.0...

Question

In a simple AC circuit shown on the right, C = 0.085 F, L = 1.9 H, R = 39 O, and I = I_maxsin(ωt), where I_max = 4.5 A and ω = 65 rad/s.

Part a. Express the capacitive reactance X_C, in terms of C, and ω.

Part b. Calculate the numerical value of X_C in ohms

Part c. Express the inductive reactance, X_L, in terms of L, and ω.

Part d. Calculate the numerical value of X_L in ohms.

Part e. Express the maximum potential difference between a and b, ΔV_ab,max, in terms of I_max and X_C.

Part f. Calculate the numerical value of ΔV_ab,maxin volts.

Part g. Express the maximum potential difference between b and c, ΔV_bc,max, in terms of I_ma_x and R.

Part h. Calculate the numerical value of ΔV_bc,maxin volts.

Part i. Express the maximum potential difference between c and d, ΔV_cd,max, in terms of I_max and X_L

Part j. Calculate the numerical value of ΔV_cd,max in volts

Patrick Ford · Accepted Answer

In this problem, we are going to state the expressions relating to quantities in RLC circuits and calculate their numerical values.Part a.We are required to give the expression that relates the capacitive reactance XC, C, and ω.[readmore]XC = 1/(ωC)The capacitive reactance XC, is equal to 1/(ωC)Part b.XC = 1/(ωC) = 1/((65)(0.085)) = 0.181ΩThe numerical value of XC is 0.181ΩPart c.Inductive reactance is expressed as:XL = ωLThe inductive reactance, XL, in terms of L, and ω is equal to ωL.Part d.We already have:ω = 65 rad/sL = 1.9 HTherefore, XL = ωL = (65)(1.9) = 123.5 ΩThe numerical value of XL is 123.5 Ω.Part e.The maximum potential difference between points a and b is:ΔVab,max = ImaxXCThe maximum potential difference between a and b, ΔVab,max, in terms of Imax and XC is ΔVab,max = ImaxXC. Part f.We're given Imax = 4.5 ABut we have XC = 0.181Ω from part b.Therefore, ΔVab,max = (4.5)(0.181) = 0.8145 VThe numerical value of ΔVab,max is 0.8145 V.Part g.The  maximum potential difference between b and c in terms of Imax and R is:ΔVbc,max = ImaxRThe expression of the maximum potential difference between b and c, ΔVab,max, in terms of Imax and R is ΔVbc,max = ImaxR. Part h.We're given Imax =4.5 Aand R = 39ΩThus, ΔVbc,max = (4.5)(39) = 175.5 VThe numerical value of ΔVbc,max in volts is 175.5 V. Part i.The maximum potential difference between c and d:ΔVcd,max = ImaxXLThe expression of the maximum potential difference between c and d, ΔVcd,max, in terms of Imax and XL is ΔVcd,max = ImaxXL.Part j.We have, from part c, XL = 123.5 ΩWe're given Imax = 4.5ASo, ΔVcd,max = (4.5)(123.5) = 555.75 VThe numerical value of ΔVcd,max in volts is 555.75 V.

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