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A load draws 10 A of current from a 120-V source. The voltage leads the currrent by...
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For the circuit to have resistive, reactive and total current it needs to be parallel (same voltage drop on all components). See figure below for a diagram of the currents in the circuit.
We know `I_T = 10 A` , `U =120 V` ,`theta =36.9 deg`
From the triangle of currents (see figure) we have:
Resistive current value
`I_R =I_T*cos(theta) =10*cos(36.9) =7.997 A =8 A`
Reactive current value
`I_X = sqrt(I_T^2 - I_R^2) =sqrt(10^2 -8^2) = 6 A`
Since the value of voltage drop U is common to all components, for the powers we have:
Real power (or equivalent Resitive Power) (notation P)
`P = U*I_R =120*8 = 960 W`
Reactive power (or equivalent Imaginary Power) (notation Pr)
`P_r = U*I_X =120*6 = 720 VAR`
Aparent power (or equivalent Total Power) (notation Pa)
`P_a = U*I_T =120*10 =1200 VA`
The triangle of powers is equivalent to the triangle of currents. For power the following relation is true:
`P_a^2 = P^2 +P_r^2`
for the absolute values of the power or written as vectors sum
Pa= P+ Pr
Posted by valentin68 on September 10, 2013 at 8:43 AM (Answer #1)
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