Question

voltmeter V1 reads 600 V, Voltmeter V2 reads 580 V, and ammeter A reads 100 A. All the measuring devices are ideal. The power wasted in the transmission line connecting the power house to the consumer is

Answer 1

Power from power house can be given as:
P1=IV1
P1=100X600=60000WATT
Power delivered to the consumer is:
P2=IV2
P2=100X580=58000WATT
WASTE OF POWER:
P=P1-P2
=60000-58000W=2000W
WASTED POWER=2000W

Answer 2

Concept:

The wastage of power can be calculated by the formula, Power = Potential difference × Current i.e. P = VI

Given:

V1 = 600 V

V2 = 580 V

Current, I = 100 A

Find:

We need to determine the power that is wasted in the transmission line connecting the powerhouse to the consumer.

Solution:

The quantity of energy moved or converted per unit of time is known as power in physics.

The watt, or one joule per second, is the unit of power in the International System of Units.

The voltage across the transmission line is given as, V = V1 - V2

Thus, V = 600 - 580

V = 20V

It has been given to us that I = 100 A

We have the formula for power loss as, P = VI where V is the potential difference while I is the current.

Therefore, Power loss = VI becomes-

P = 20 × 100

P = 2000 W

P = 2kW

Thus, The power wasted in the transmission line connecting the powerhouse to the consumer is 2kW.

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