Question

: Referring to Problem 2.5, determine the instantaneous power, real power, and reactive power absorbed by (a) the 20-Ω resistor, (b) the 10-mH inductor, (c) the capacitor with 25-Ω reactance. Also determine the source power factor and state whether lagging or leading.

Short answer

(a) Therefore, the expression for the instantaneous power is .

p (t) = 2000 [cos 92$\mathrm{\omega }$t - 50^o) +0.342^o] . The value of the real power is .

3.385kW. The value of the reactive power is 0 and the value of power factor is 1.

(b) Therefore, the expression for the instantaneous power is .

p (t) = (2.035x10⁴)W The value of the real power is 0 . The value of the reactive power is 20.35kVAR and the value of power factor is 0 lagging.

(c) Therefore, the expression for the instantaneous power is .

p 9t) = (3.069x10³) cos (2 wt +150^o)WThe value of the real power is 0 . The value of the reactive power is -3.069kVAR and the value of power factor is 0 leading .

Step-by-step solution

Define formula for the instantaneous power, real power reactive power and the active power.

Consider the expression for the instantaneous current.

i (t) = V (t) / R .............(1)

Consider the formula for the average power absorbed by the element.

p (t) = v (t) i (t) ..............(2)

Here, v (t) is the instantaneous voltage and i (t) is the instantaneous current.

Consider the formula for the real power.

..............(3)

Here, θ is the phase angle, V_ms and l_msare the rms voltage and current respectively.

Consider the formula for the reactive power.

..............(4)

Consider the formula for the power factor.

PF = 2π fL ..............(5)

Consider the formula for the inductive reactance.

X_L= 2πfL ...................(6)

Write the formula for the current through the inductor.

.............(7)

Write the formula for the current through the capacitor.

..........(8)

Determine the expression for the instantaneous voltage and reactance.

Consider the formula to determine the instantaneous voltage.

v (t) = v_maxcos ($\mathrm{\omega }$t +$\mathrm{\delta }$)

Here, V_max is the maximum voltage and is given as,

V_max = $\sqrt[]{2\mathrm{V}}$_ms

Since, the source voltage is V =277 ∠ 30^oV then, the rms voltage is V _ms= 277V,

Solve for the instantaneous voltage as,

Substitute values into equation (6) and solve for inductive reactance.

Solve for instantaneous power, active power, real power, reactive power and power factor for 20-Ω resistance.

Substitute the values in equation (1) and solve for the instantaneous current.

Substitute the values in equation (2) and solve for the instantaneous power.

p (t) =391.7cos ($\mathrm{\omega }$t +30^o) 19.58cos ( $\mathrm{\omega }$t +30^o)

Solve further as,

Note that resistance neither absorb nor supply reactive power, so the reactive power is,

Substitute the values in equation (3) and solve for the real power.

Substitute the values in equation (4) and solve for the reactive power.

Substitute the value in equation (5) and solve for the power factor.

PF = cos (0) =1

Therefore, the expression for the instantaneous power is

p (t) = 2000 [cos 92ωt - 50^o) +0.342^o] . The value of the real power is 3.835kW . The value of the reactive power is 0 and the value of power factor is 1.

Solve for instantaneous power, active power, real power, reactive power and power factor for 10-mH inductor

Substitute the values in equation (1) and solve for the instantaneous current.

Substitute the values in equation (2) and solve for the instantaneous power.

Consider the phase angle is the difference between the voltage and the current angle as,

Substitute the values in equation (3) and solve for the real power.

Substitute the values in equation (4) and solve for the reactive power.

Substitute the value in equation (5) and solve for the power factor.

PF = cos (90) =0

Therefore, the expression for the instantaneous power is

p (t) = (2.035x10⁴)cos (2ωt-30^o)W The value of the real power is . The value of the reactive power is 20.35kVAR and the value of power factor is 0 lagging.

Solve for instantaneous power, active power, real power, reactive power and power factor for 10-mH inductor.

Substitute the values in equation (1) and solve for the instantaneous current.

Substitute the values in equation (2) and solve for the instantaneous power.

Consider the phase angle is the difference between the voltage and the current angle as,

Substitute the values in equation (3) and solve for the real power.

Substitute the values in equation (4) and solve for the reactive power.

Substitute the value in equation (5) and solve for the power factor

PF = cos (-90)

=0 leading

Therefore, the expression for the instantaneous power is .p (t) = (3.069x10³)cos (2ωt-150^o)W The value of the real power is . The value of the reactive power is -3.069kVAR and the value of power factor is 0 leading .