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dangina [55]
4 years ago
5

A power station delivers 510 kw of power at 12,000 v to a factory through wires of total resistance of 2.5 ω. how much less powe

r is wasted if the electricity is delivered at 47,000 v rather than 12,000 v?
Physics
1 answer:
bixtya [17]4 years ago
3 0

The currents (Amperes) in the two cases are calculated as power/voltage: <span>

510000W / 12000V = 42.5 A 
510000W / 47000V = 10.85 A 

<span>And that the power wasted in the two cases is calculated as current^2 x resistance:

(42.5 A)^2 (2.5 Ohms) = 4,515.625 W 
(10.85 A)^2 (2.5 Ohms) = 294.30625 W 

<span>So the amount of power saved is the difference:
<span>4,515.625 - 294.30625  = 4221.32 Watts</span></span></span></span>

<span>This is the main reason why distribution grids aim to operate at the highest voltages possible.</span>

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3.1
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Answer:

The value is  C = 30729\  c

Explanation:

From the question we are told that

  The power rating of the stove is  P = 4.4 KW  = 4.4 *10^{3} \  W

   The duration of its use everyday is t_1  = 70 \ minutes  = 1.167 \ hours    

    The rating of the light bulbs is P_2'  = 150 W

   The number is n = 7

   The power rating of the total  bulb is  P_2 = 7 * 150 = 1050 \  W

    The duration of its use everyday is  t_2  = 7 hours

    The power rating of miscellaneous appliance P_3 = 1.8 \ KW = 1.8 *10^{3} \  W

    The duration of its use everyday is t_3 = 1 hour

     The power rating of hot water P = 4 KW  = 4 *10^{3} \  W

      The duration of its use everyday is t_4 = 120 \ minutes  = 2 hours

Generally the total electrical energy used in 1 month is mathematically represented as

    E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30

=>  E = 4.4*10^{3} * 1.167  * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30

=>  E = 614598  \ W \cdot h

=>   E = 614.6  \  K W \cdot h

Generally the monthly electricity  bill is mathematically represented as

    C = 614.6 * 50

=> C = 30729\  c

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