Missing a Phase Measurement

Electricity and Magnetism Level 3

A three-phase A C AC load with no neutral connection (and correspondingly no neutral current) draws currents on the A A and C C phases, as shown (RMS values). The B B -phase current is not measured.

All three load resistances are each 1 Ω 1\, \Omega .

How much power does the three-phase load draw in total (in watts)?


The answer is 4.

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Oct 6, 2017

There is no neutral path, so the sum of the three currents must be zero.

I A ⃗ + I B ⃗ + I C ⃗ = 0 I B ⃗ = − ( I A ⃗ + I C ⃗ ) = − ( 1 + j 0 + 0 + j 1 ) = − 2 ∠ 4 5 ∘ = 2 ∠ − 13 5 ∘ \vec{I_A} + \vec{I_B} + \vec{I_C} = 0 \\ \vec{I_B} = -(\vec{I_A} + \vec{I_C}) = -(1 + j0 + 0 + j1) = -\sqrt{2} \angle 45^\circ = \sqrt{2} \angle {-135^\circ}

The total power consumed by the load is:

P t o t a l = ∣ I A ⃗ ∣ 2 R A + ∣ I B ⃗ ∣ 2 R B + ∣ I C ⃗ ∣ 2 R C = 1 2 ( 1 ) + ( 2 ) 2 ( 1 ) + 1 2 ( 1 ) = 4 P_{total} = |\vec{I_A}|^2 R_A + |\vec{I_B}|^2 R_B + |\vec{I_C}|^2 R_C \\ =1^2 (1) + (\sqrt{2})^2 (1) + 1^2 (1) = \boxed{4}

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