A load draws 10 A of current from a 120-V source. The voltage leads the currrent by 36.9 degrees. How do I determine the resistive current, reactive current, and power used by the load?

1 Answer | Add Yours

Top Answer

valentin68's profile pic

Posted on

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`

Observation:

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

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Sources:

We’ve answered 324,723 questions. We can answer yours, too.

Ask a question