In Blondel's theorem, the total power in a network of N conductors is measured by how many watt-meters?

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Multiple Choice

In Blondel's theorem, the total power in a network of N conductors is measured by how many watt-meters?

The total active power in a network with N conductors can be obtained by adding up the power contributed by each conductor. Blondel's theorem states that you can do this by using one wattmeter for every conductor, so the total power is the sum of the readings from all N wattmeters. Intuitively, each conductor carries a voltage with respect to a common reference and has a current flowing through it. If a wattmeter is connected so it measures the product of that conductor’s voltage and its current (v times i), its reading represents the power associated with that conductor. Adding these powers for all conductors gives the total active power of the network. In a balanced three-phase setup, you might sometimes measure total power with fewer meters due to symmetry, but Blondel’s general statement for a network with N conductors is that N wattmeters are sufficient.

The total active power in a network with N conductors can be obtained by adding up the power contributed by each conductor. Blondel's theorem states that you can do this by using one wattmeter for every conductor, so the total power is the sum of the readings from all N wattmeters.

Intuitively, each conductor carries a voltage with respect to a common reference and has a current flowing through it. If a wattmeter is connected so it measures the product of that conductor’s voltage and its current (v times i), its reading represents the power associated with that conductor. Adding these powers for all conductors gives the total active power of the network.

In a balanced three-phase setup, you might sometimes measure total power with fewer meters due to symmetry, but Blondel’s general statement for a network with N conductors is that N wattmeters are sufficient.

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