Question

please help, if you're able to explain a-f that would be great, but i mainly need help with G

<Ch31. AC Circuits Part Part A Exercise 31.12 - Enhanced - with Feedback What is the impedance of the circuit? Express your answer with the appropriate units. What is the voltage amplitude across the resistor? Express your answer with the appropriate units. 2 = 251 VR = 19.7V Previous Answers Sube Previous Answers You have a 170 resistor, a 0.800 H inductor, and a 7.40 uF capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 29.0V and an angular frequency of 230 rad/s. ✓ Correct ✓ Correct Part B Part D Part E What is the current amplitude? Express your answer with the appropriate units. What is the voltage amplitude across the inductor? Express your answer with the appropriate units. What is the phase angle of the source voltage with respect to the current? Expre your answer in degrees T = 0.116 A V.- 213 V 4 47.3 Previous AS Previous Answers Previous Aws Correct ✓ Correct ✓ Correct Part F Part 6 Does the source vatage lag er lead the curent? Construct the phasor diagram. Draw the phasors starting at the black dot. The location and orientation of the phasors will be graded. The length of the phasors will not be graded. The voitage as the current The voltage leads the current Press Correct Vectors: V Resistor voltage phasor V Inductor voltage phasor

Answer 1

(31.12) Given data;

Resistance Indutance and capacitance is given as;

R = 17022, L=0.8 H, and C= 7.4 x 10-6 F

source voltage E = 29 V;

angular frequency w = 230 rad/s;

Solution:

(A) Impedence of the circuit;

R2 + (XL - Xc)'

since we use only resistor and inductor in series thus Xc = 0

R2 + (WL)?

1702 + (230 x (230 x 0.8)

2 = 251 Ω

(B) by using ohm's law current amplitude is defined as;

Ε 1

29 1 250.5

= 0.116 A

(C) voltage amplitude across the resistor

VR IR= 0.116 x 170

VR = 19.7 V

(D) voltage amplitudes across the inductor

IXL IWL 0.116 x 230 x 0.8

VI = 21.3 V

(E) Phase angle;

-1 tan = ي L R

-1 0 = tan 230 x 0.8 170

o = 47.39

(F) since, phase angle is positive, so voltage source is leading the current.

(G) Phasor diagram:

XL Z 47.3 → R