urgent need of hints and formulae! to calculate a)current b) power c)the power factor?

two coils are connected in series with 2 A d.c through the circuit , the p.ds across the coils are 20V and 30V respectively. With 2 A a.c at 40 Hz, the p.ds across the across the coils are 140V and 100V respectively . If two coils are connected to 230 V , 50 Hz supply.

it will be great if having the solution ...

thanks

Erica

Update:

mr uncouth your answers are perfect but i didn't get this

So (X of L) at 50Hz = 5/4 (X of L) at 40 HZ.

Therefore at 50Hz:

(X of L1) = 5/4 X ( 69.28 Ohm) = 86.6 Ohm

(X of L2) = 5/4 X (47.70 Ohm) = 59.625

i mean how it can be .. can someone explain?? i will thankful !

2 Answers

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  • 1 decade ago
    Favorite Answer

    I am not familiar with the meaning of "p.ds" so my answers may be incorrect but the following should give you an idea on how to solve all parts of the problem. Also, I assumed a sine wave source Voltage.

    R of L1 = 20V / 2A = 10 Ohm at any frequency

    R of L2 = 30 V / 2A = 15 Ohm at any frequency

    At 40Hz:

    Z of L1 = 140V / 2A = 70 Ohm

    X of L1 = sq rt of (70 Squared - 10 squared) = 69.28 Ohm

    Z of L2 = 100V / 2A = 50 Ohm

    X of L2 = sq rt of (50 squared - 15 squared) = 47.70 Ohm

    (X of L) is directly proportional to frequency. That is (X of L) = 2pi f L

    So (X of L) at 50Hz = 5/4 (X of L) at 40 HZ.

    Therefore at 50Hz:

    (X of L1) = 5/4 X ( 69.28 Ohm) = 86.6 Ohm

    (X of L2) = 5/4 X (47.70 Ohm) = 59.625

    Since the coils are in series total inductive reactance = (86.6 + 59.625) Ohms = 146.225 Ohms and

    The total resistance = (R of L1) + (R of L2) = (10 + 15) Ohms = 25 Ohm

    The total impedance = Zt = sq rt of [(25squared) + ( 146.225squared)] = 148.35 Ohm

    Answers:

    (a) current = Volts / total impedance = 230V / 148.35 Ohm = 1.55 Amps

    (b) Power = (current squared) X (resistance) = (1.55 Amps) squared x (25 Ohm) = 60 Watts

    (c) power factor = (total resistance) / (total impedance) = (25 Ohms) / (148.35 Ohms) = .168

    Also power factor = (Watts) / (Volt-Amps) = 60W / (230V X 1.55A) = .168

    Response to additional details:

    X of L = 2pi f L = (6.28) X (frequency) X (Inductance).

    I did not use this formula to calculate (X of L) at 40 HZ, because I did not know the value of L. But since nothing else changed but the frequency, that allows me to use the ratio of the two frequencies and the value of (X of L) at 40Hz to calculate (X of L) at 50Hz by simply multiplying (50Hz / 40Hz) or (5/4) X (X of L) at 40 Hz. Look at the formula (X of L) = 2pi f L and you will surely see why this is possible.

  • Anonymous
    1 decade ago

    (1) in dc R values are obtained

    R1 = 20/2=10ohms

    R2 = 30/2=15

    (2) in ac Z values are obtained

    Z1 = 140/2=120

    Z2 = 100/2=50

    (3) now you have the two coils across 230v ac

    here are the formulae

    draw a right triangle

    horiz line is R or W

    vert line is XL or VAR

    the slope is Z or VA

    let angle between slope and horiz be Q

    pf = cos(Q)

    pf = R/Z

    pf = W/VA

    then some more formulae

    Z = sqrt[R^2 + XL^2]

    complex number for RL

    R + XL j

    addition of series RL

    (R1+R2) + (XL1+XL2) j

    note that frequency is not used unless you are dealing with inductance L

    XL = 2pi f L

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