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1 large room is illuminated by twenty 150-watt lamps and thirty 100-watt lamps. If the circuit voltage is 116, calculate the total current. Given: (20x) P 1 = 150W (30x) P 2 = 100W E = 116V Required: I =? Solution: P 1 + P 2 = PT P 1 + P 2 = IE 20 (150W) + 30 (100W) = I (116V) I = 51 Calculate the r...

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1.A large room is illuminated by twenty 150-watt lamps and thirty 100-watt lamps. If the circuit voltage is 116, calculate the total current. Given: (20x) P1 = 150W (30x) P2 = 100W E = 116V Required: I=? Solution: P1 + P2 = PT P1 + P2 = IE 20 (150W) + 30 (100W) = I (116V) I = 51.72A 2. Calculate the resistance of a load that takes 1,600-watts from a 220-volt source. Given: P = 1600W E = 220V Required: R=? Solution: R = V2/P R = (220V)2/(1600W) R = 30.25|

3. How much power is represented by a circuit in which the voltage and current equations are e = 150 sin 314t and i = 42.5 sin 314t?

Given: I = 42.5 sin 314t E = 150 sin 314t Required: P=? Solution: P = EmIm/2 P = (150V) (42.5A)/2 P = 3400W 4. A coil of wire having negligible resistance and and inductance of 0.248-henry is connected to a 117-volt 50-cycle source. Calculate (a) the inductive reactance, (b) the current, (c) the maximum power delivered to the inductor or returned to the source, (d) the average power. Write the equations for (e) the current and (f) the power. Given: L = 0.248H E = 117V f = 50Hz Required: XL, I, Pm, Pave, equation of I & P Solution: a.) XL = 2πfL = 2π (50Hz) (0.248H) = 77.91| b.) I = E/XL = 117V/ 77.91| = 1.5A c.) Pm = EI = 117V (1.5A) = 175.5W d.)

θ

= sin-1(XL/Z) = sin-1(77.91/ 77.91) = 90°

Pave = EIcos

θ

= 117V (1.5A) cos (90°) = 0

e.) Im =1.5A √ 2 = 2.12A I = 2.12sin[(2π)(50)t-90(π/180)] I = 2.12sin (314t- π/2) f.) P = Pmsin2t P = - 175.5 sin [(2)(314)t] P = - 175.5 sin 628t 5. At what frequency will a 0.016-henry inductor have an inductive reactance of 80-ohms? Given: L = 0.016H XL = 80| Required: f=? Solution: XL = 2πfL 80| = 2πf(0.016H) f = 795.77 Hz 6. A 0.15-henry inductor is connected to a 60-cycle source of emf whose rms value is 113-volts. How much energy is stored in the magnetic field (a) during a period when the current rises from zero to its maximum value?, (b) during the period represented by a positive half-cycle of voltage plus half of the succeeding negative half-cycle of voltage? Given: L = 0.15H

f = 60Hz E = 113V Required: W Solution: a.) W = L I2 W = 15H

(

113V 2 π (60Hz)(0.15H)

)

2

= 60joules

b.) W = L I2 W = - 15H

(

113V 2 π (60Hz)(0.15H)

)

2

= - 60joules

7. What value of capacitance will have a capacitive reactance of 80-ohms at 796-cycles? Given: f = 796 Hz Xc = 80| Required: C=? Solution: Xc = 1/(2πfC) 80| = 1/[2π (796Hz)(C)] C = 2.5µF 8. A 45-µF capacitor is connected to a 118-volt 60cycles source. Calculate (a) the capacitive reactance, (b) the current, (c) the maximum power

delivered to the capacitor or returned to the source, (d) the average power. Write equations for (e) the current and (f) the power. Given: C = 45µF E = 118V f = 60Hz Required: XC, I, Pm, Pave, equation of I & P Solution: a.) XC = 1/2πfL = 1/2π (60Hz) (45µF) = 58.95| b.) I = E/XC = 118V/ 58.95| = 2A c.) Pm = EI = 118V (2A) = 236.22W d.)

θ

= sin-1(XL/Z) = sin-1(58.95/58.95) = 90°

Pave = EIcos

θ

= 118V (2A) cos (90°) = 0

e.) Im =2A √ 2 = 2.83A I = 2.83sin[(2π)(60)t+90(π/180)] I = 2.83sin (377t+ π/2) f.) P = Pmsin2t P = 236.22 sin [(2)(377)t] P = 236.22 sin 754t

9. What value of capacitance will have a capacitive reactance on 180-cycles that is equal to the 60cycle inductive reactance of a 0.0061-henry inductor?

Given: f1 = 180Hz f2 = 60Hz L = 0.0061H Required: C=? Solution: XL = 2π(180Hz)(0.0061H XL = 23| XL = XC 23| = 1/[2π(60Hz)(C)] C = 38.45µF 10. A sinusoidal emf represented by the equation e = 340 sin 314t is impressed across a 25-µF capacitor. How much energy is stored in the electric field on 0.025 sec? (Start the timing period as the emf wave passes through zero.) Given: e = 340 sin 314t C = 25µF T = 0.025 sec Required: W=? Solution: W = CE2 = (25x10-6F)(340/ √ 2 )2 = 1.445 joules

11. The equation of the current in a pure inductor circuit is i = 4.25 sin 157t when the impressed emf wave is e = 650 cos 157t. Determine the equation of the power wave. Given: I = 4.25sin157t E = 650cos157t Required: Equation of P Solution: P = Irms Vrms sin 2ѡt P = [(-4.25) (650)/2]sin2(157)t P = -1381.25sin314t 12. The current in a pure capacitor circuit is given by the equation i = 0.17 cos 5000t. If the rms value of the impressed emf is 80-volts, (a) what are the equations for the voltage and power?, (b) what is the circuit frequency? Given: i = 0.17 cos 5000t Erms = 80V Required: a.) Equation of E & P b.) F = ? Solution: a.) Vm =80 √ 2 V = 113.14V V = 113.14sin5000t

P = Irms Vrms sin 2ѡt P = (0.17A/ √ 2 )(80V)sin2(5000)t P = 9.62sin10000t b.) Ѡ = 2πf 5000 = 2πf f = 795.77 Hz 13. At what frequency will the inductive reactance of a 0.0211-henry inductor be equal to the capacitive reactance of a 75-µF capacitor? Given: L = 0.0211H C = 75µF Required: f=? Solution: 2πfL = 1/(2πfC) 2πf(0.0211H) = 1/[2πf(75µF)] f = 126.52 Hz 14. A series circuit consisting of a 30-µF capacitor and a 0.155-henry inductor is connected to a 120volt 60-cycle source. Calculate the circuit current and indicate whether it lags behind or leads the voltage. Given: C = 30µF

L = 0.155H E = 120V f = 60 Hz Required: I=? Solution: E = IZ 120V = I [2π(60Hz)(0.155H) - 1/2π(60Hz)(30µF)] I = - 4A I = 4A leading 15. If a variable inductor is substituted for the one in Prob. 14, what should be its value if an equal current is to lag behind the voltage? Assume all other conditions to remain unchanged. Given: C = 30µF I = 4A lagging E = 120V f = 60 Hz Required: L=? Solution: E = IZ 120V = (4A)

[

( 2 π) ( 60Hz )( L ) -

1 2 π ( 60Hz)( 30 µ F )

L = 0.314H or 314.08mH

]

16. A series circuit consisting of a 0.398-henry inductor and a 212-µF capacitor is connected to a 125-volt variable-frequency source. At what frequency will the circuit take a lagging current of 2.5-amp? Given: L = 0.398H C = 212µF E = 125V I = 2.5A lagging Required: f=? Solution: E=IZ 125V=(2.5A)

[

( 2 π) ( f ) ( 0.398H ) -

1 2 π( f )( 212 µ F )

f=30Hz

17. Solve Prob. 16 for a lea Given: L = 0.398H

]

C = 212µF E = 125V I = 2.5A leading Required: f=? Solution: E=IZ 125V=(-2.5A) [

( 2 π) ( f ) ( 0.398H) -

1 2 π ( f )( 212 µ F )

]

f=10Hz

18. A 0.143-henry inductor is connected in series with a variable capacitor to a 208-volt 400-cycle source. For what value of capacitance will the current be (a) 1.04-amp lagging? (b) 1.04-amp leading? Given: L = 0.143H E = 208V f = 400Hz I = 1.04A Required: a.) C=? at 1.04A lagging b.) C=? at 1.04A leading Solution: a.) E=IZ

208V=(1.04A)

[

( 2 π) ( 400Hz ) (0.143H) -

1 2 π ( 400Hz ) (C )

]

C=2.496 µF b.) E=IZ 208V=(-1.04A) [

( 2 π) ( 400Hz ) (0.143H) -

1 2 π ( 400Hz ) (C )

]

C=0.711 µF 19. An impedance coil has a resistance of 7.5-ohm and an inductive reactance of 18-ohms. (a) What is the equation of the voltage wave that produces a current i = 11.3 sin wt? (b) Calculate the values of E, I, EE, and EL. Given: R = 7.5| XL = 18| i = 11.3 sin wt Required: a.) Equation of E b.) E, I, EE and EL Solution: a.)

( )

Im=Erms √ 2

θ = tan−1 18 7.5

Im=155.81V √ 2

θ =0.374 π

Im=220.35V E=220.sinw(wt+0.374

π

)

Im

b.) Irms= √2 =

11.3 √2

I=7.99A 2 2 E=I ( √R + X L) = (7.99A) ( √ (7.5|) + (18|) ) 2

E=155.81V EL=IXL EL=(7.99A)(18|) EL=143.83V ER=IR ER=(7.99A)(7.5|) ER=59.93V

2

20. A small a-c motor used in a washing machine is, in effect, an R-L circuit. If the machine takes 311-watts and 4.5-amp from a 115-volt source when operating normally, calculate its power factor. Given: P = 311W I = 4.5A E = 115V Required: PF = ? Solution: S = EI = (115V)(4.5A) S = 517.5 PF =

P S

=

311W 517.5W

PF = 0.601 lagging 21. A Transformer takes 5,175-watts at a power factor of 0.85 when connected to a 2,300-volt distribution circuit. What is the current input? Given: P = 5175W PF = 0.85 V = 2300V Required:

I=? Solution: PF = P/S

I = S/E

0.85 = 5175W/S

I = 6088.24/2300V

S = 6088.24

I = 2.65A

22. In a-c circuit the sinusoidal voltage and current waves have the following equations: e = 170 sin 314t, i = 28.4 sin (314t - π/3). Calculate the following: (a) effective voltage and current, (b) frequency, (c) power factor, (d) power. Given: e = 170 sin 314t i = 28.4 sin (314t - π/3) Required: E, I, f, PF, P Solution: a.) Irms = 28.4/ √ 2 Erms = 170/ √ 2

= 20. 08A

Erms = 120. 21V b.) f = 314/2π f = 49.97Hz c.)

θ

= - π/3 PF = cos θ = cos (- π/3) PF = 0.5 lagging

d.) S = IE = (20.08A)(120.21V) S = 2414V P = Scos θ = 2414V(0.5)

P = 1207W 23. An impedance coil has a resistance of 20-ohms and an inductive reactance of 40-ohms. For what value of series resistance will the overall power factor of the circuit be 0.8? Given: R = 20| XL = 40| PF = 0.6 Required: R=? Solution: Z2 = (R + 20)2 + (40)2 R + 20

0.6 = √(R+20) + (40) 2

2

R = 10| 24. A series R-L circuit takes 371.2 watts at a power factor of 0.8 from a 116-volt 60-cycle source. What are the values of R and L? Given: P = 371.2W PF = 0.8 E = 116V f = 60Hz Required: R, L...