Calculate the reading of each wattmeter in the circuit shown below if the phase voltage at load
is 120 V and Z ϕ = 10∠-75° Ω. Show that the sum of the wattmeter readings equals the total
average power delivered to the load.
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Answer:
P_ T (d)=3(12)² (2.5882)=1118.10W,
W 1 +W 2 =1763.63–645.53=1118.10W.
Step-by-step explanation:
Z ϕ =10∠−75 ∘ Ω,
V_{L}=120\sqrt{3}VV
L =120\sqrt{3} V, and I_{L}=12AI
L =12A.
W_{1}=\left(120\sqrt{3}\right)(12)\cos\left(-75^{\circ}+30^{\circ}\right)= 1763.63
W1 =(120\sqrt{3} )(12)cos(−75° +30 °)=1763.63W,
W2 =\left(120\sqrt{3}\right)(12)\cos\left(-75^{\circ}-30^{\circ}\right)= -645.53
W2=(120 3 )(12)cos(−75 °−30° )=−645.53W.
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