6, Aseries LCR circuit with L-160 mH . C·100 F. and R-400 is connected to a sinu

Question

6, Aseries LCR circuit with L-160 mH . C·100 F. and R-400 is connected to a sinusoidal voltage V (t)- (40.0V)sin(wt), with w-200 tad/sLet the eurrent at any instant in the eireuit be I()lo sinut-).Find l? (a) 2.121 A moves with a constant n. It enters a reogion (e) 0.854 A B directed along the (b) 0.143 A (d) 0.911 A 7. Aseris LCR circuit with L =100 mll . C-100," F, and R=40on is connected to a sinusoidal voltage V (t) the particle can just (400v7min(wt) , with w = 200 rad/s, what is the phase constant ? (a) -54.3 (e)24.20 (b)-24.2 (d)-84.3 vertically downwards wards you and enters when seen from above 8. Suppose an AC generator with V (t) (150V)sin(100t) is connected to a series LCR circuit with R-40.00, L = 800 mH , and C00 F, shown in figure. Calculate the maximum voltage across the resistor (VRo) rtical anticlockwise rizontal anticlockwise charged parallel plate The space between e field of induction B. vn-Fe sinon on in the capacitor is: (a) 20.5 V (c) 56.3 V 9. Suppose an AC generator with V (t)-(150V)sin(100t) is connected to a series LCR circuit with R-400 , L = 80.0 mH , and C = 50.0 F, as shown in figure (qu'etion 8). Calculate the maximum voltage across the inductor (VLo)? (b) 14.2 V (d) 30.6 V lerated through a (a) 11.52 V ar magnetie field (c) 6.31 V eased to 5V, the 10. Suppose an AC generator with V (t) = (150)sin(100) (b) 4.12 V (d)6.12 v is connected to a series LCR circuit with R = 40.0 , L = 80.0 mH , and C = 50.0 ALF. Calculate the maximum potential difference across the inductor and the capacitor between points b and d shown in figure (question 8)? 100 F, (b) 140 V oltage V (t) # (a) 120 V edance of the (c) 147 V (d) 138 V

Explanation / Answer

7)Correct Ans is:

(b)-24.2 deg

explanation:

Given,

L = 160 mH ; C = 100 uF ; R = 40 Ohm

Vm = 40 V ; w = 200 rad/s

we know that the phase angle is given by

phi = tan^-1 [(XL - Xc)/R]

XL - Xc = w L - 1/wc

XL - Xc = 200 x 160 x 10^-3 - 1/200 x 100 x 10^-6 = -18

theta = tan^-1(-18/40) = -24.23 deg

Hence, (b)theta = -24.2 deg

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