# 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

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

**Sources:**