CPHY 122: PHYSICS 2 Electricity and Magnetism TEST III, Due Friday, April 13, 20

Question

CPHY 122: PHYSICS 2 Electricity and Magnetism TEST III, Due Friday, April 13, 2018, 5:30 PM er ALL questions. Show clearly your calculations for fall marks 1(a) A seties ac circuit contains the following components R-150 ?, L-250 mil, C-2 ?F and a generator with V- 210 V operating at 50 Hz. Calculate the (a) inductive reactance, (b) capacitive reactance (c) impedance, (d) peak current, and (e) phase angle. I(b) A sinusoidal voltage e(c)-(40 V) sin(100) is applied to a series RLC circuit with L-160 mil, C-99 ?F and R-68 ?. (a) What is the impedance of the cir. cuit? (b) What is the current amplitude? (c) Deter- mine the numerical values for·?, and ? in the . equation io)- sinlost- 2. An 80-2 resistor, a 200-mH inductor, and a 0.15-4F cenacitou tare conected in paralel ros s iga de (ems) source operating at an angular frequency of 374 rad/s. (a) What is the resonant frequency of the circuit? (b) Calculate the rms current in the resistor, inductor, and capacitor. (c) What is the rms current delivered by the source? (d) Is the current leading or lagging behind the voltage? By what angle? 3. A 1000-? resistor is connected in series to a 0.6-H inductor and a 2.5-uF capacitor. This RLC combina tion is then connected across a voltage source that Caleulate (a) the varies as (80 V) sin1000 peak current, (b) the phase angle, (c) the power fac tor, (d) V across the inductor, and (e) the average power delivered to the circuit. In the cireuit shown in Fig. 31-66, 10V.R 5.00, Rs-10 ?. and 1-5.0 H. For two separase conditions (1) witch S just closed and () swich S closed for a long time calculate (a) the current i, through R?(b) the current 4 through R3. (c) the current i through the swiach, (4) the potential f fereace across R. () the potential difference across L, and (f) the rate of change dildt 4. A long straight wire is parallel to one edge and is in the plane of a single turn rectangular loop as in Figure 31.44. (a)If the current in the long wire varies in time as i -Ie-, show that the induced emfin the loop i b) Calculate the value for the induced emf att-5 taking I,- 10 A, d 3 em,a 6em, b-15 am, and FIGURE 31-66 Figure 3144 In an LCR circuit, R- 16.0n, C- 31.2 F, L- 9.20 mH, aed 1-1-smcar, with C.-45 V, and ?- 3000 rad/s. For time r 0.442 ms find (a) the rate at which energy is being supplied by the generator, (6) the rate at which being stored in the capacitor, (c) the rute a which energy is being stored in the ieducior, and (d) the rate at which energy is being dissipated in the resistor (e) What is the 6. meaning of a negative result for any of parts (a), (bla nd eg U) Show that the results of parts (b) (e), and (d) sum to the result of part (a).

Explanation / Answer

1 a. a series AC circuit contains

R = 150 ohms

L = 250 mH

C = 2 uF

Vm = 210 V

f = 50 Hz

a. inductive reactance = Lw = 2*pi*L*f = 2*pi*250*50/1000 = 78.539816339744830 ohms

b. capacitative reactance, 1/Cw = 1/2*pi*f*C = 1591.549430 ohms

c. impedance = sqrt(R^2 + (Xl - Xc)^2) = 1520.42694366 ohms

d. peak current Im = Vm/Z = 0.1381190992929 A

e. phase angle = arctan((Xl - Xc)/R) = -84.338189 deg

2. R = 80 ohm

C = 0.15 uF

connected in parallel across

V = 120 V rms

w = 374 rad/s

L = 200 mH

a. resonant frequency = 1/sqrt(LC) = 5773.502691 rad/s

fr = 918.881492369653 Hz

b. rms current in resistor = Vrms/R = 1.5 A

rms current in capacitor = Vrms/Xc = Vrms*w*C = 0.011781 A

rms current in inductor = Vrms/Xl = Vrms/Lw = 2.80748663101 A

c. rms current delivered by the source = V'

now, V' = |1.5 + 2.80748663101j - 0.011781 j|

V' = sqrt(1.5^2 + 2.79570563101^2) = 3.172691282690 A

d. current is lagging the voltage ( as capacitative reactance is more than inductive reactance)

angle = arctan((Xl - Xc)/R) = -89.7417746 deg

3. R = 1000 ohm

L = 0.6 H

C = 2.5 uF

series circuit

V = 80 sin(1000t/pi)

a. w = 1000/pi

hence

Z = sqrt(R^2 + (wL - 1/wC)^2) = 1461.3734397085 ohm

ip = Vp/Z = 80/Z = 54.74302312210 mA

b. phase angle = arctan((Xl - Xc)/R) = -46.82037849 deg

c. power factor = cos(phi) = 0.6842877890

d. Vrms acxross inductor = ip*Xl/sqrt(2) = 7.39290553688 V

e. average power = Vp*Ip*cos(phi)/2 = Vp^2*cos(phi)/2Z = 1.49839929021618416 W

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