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28 December, 14:48

Your car's 30.5 W headlight and 2.75 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would one headlight and the starter consume if connected in series to a 12.0 V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices. Answer to the nearest 0.1 W.)

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Answers (2)
  1. 28 December, 14:57
    0
    Given Information:

    Headlight power = Ph = 30.5 W

    Starter power = Ps = 2.75 kW = 2750 W

    Voltage = V = 12 V

    Required Information:

    Pseries = ?

    Answer:

    Pseries = 30.2 W

    Explanation:

    First we have to find the resistance of headlight and starter

    As we know power is given by

    P = V²/R

    For headlight:

    Ph = V²/Rh

    30.5 = 12²/Rh

    Rh = 12²/30.5

    Rh = 4.72 Ω

    For starter:

    Ps = V²/Rs

    2750 = 12²/Rs

    Rs = 12²/2750

    Rh = 0.0523 Ω

    In a series connection, the equivalent resistance will be

    Req = Rh + Rs

    Req = 4.72 + 0.0523

    Req = 4.772 Ω

    Therefore, the power consumed by headlight and the starter is

    Pseries = V²/Req

    Pseries = 12²/4.772

    Pseries = 30.18

    Pseries = 30.2 W
  2. 28 December, 15:09
    0
    Headlight, P = 29.7W and starter P = 0.330W.

    Explanation:

    Given V = 12V

    P = I*V = V/R * V = V²/R

    So the resistance for each element is given by R = V²/P

    So for the head light,

    P = 30.5W,

    R1 = 12²/30.5 = 4.72Ω

    For the starter P = 2.75kW = 2750W

    R2 = 12²/2750 = 0.0524Ω

    Now in series connection the same current I flows through them.

    So Req = R1 + R2 = 4.72 + 0.0524 = 4.7724Ω

    I = V/Req = 12/4.7724 = 2.51A

    So the power consumed by:

    The headlight is P = I²R1 = 2.51²*4.72 = 29.7W.

    The Starter P = I²*R2 = 2.51²*0.0524 = 0.330W
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