Question

The voltage v (t) =359.3cos (ωt) volts is applied to a load consisting of a 10- resistor in parallel with a capacitive reactance X_c 25Ω . Calculate (a) the instantaneous power absorbed by the resistor, (b) the instantaneous power absorbed by the capacitor, (c) the real power absorbed by the resistor, (d) the reactive power delivered by the capacitor, and (e) the load power factor.

Short answer

(a)Therefore, the instantaneous power that is absorbed by the resistor is

6455+6455cos (ωt) W

(b)Therefore, the instantaneous power that is absorbed by the capacitor is

-2582sin (ωt) W

(c)Therefore, the real power absorbed by the resistor is 6455 W .

(d)Therefore, the real power delivered by the capacitor is 2582 VAR .

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) = $\frac{\mathrm{V}\left(\mathrm{t}\right)}{\mathrm{R}}$ ...........(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.

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

Here, θ is the phase angle, V_rms and l_rmsare the rms voltage and current respectively.

Consider the formula for the reactive power.

$Q=\frac{V}{\sqrt{2}}\frac{I}{\sqrt{2}}\sin \left(\varphi \right)$ ..........(4)

Determine the expression for the instantaneous power absorbed by the resistor.

(a)

Note that the capacitor and the resistor are connected in parallel.

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

$\begin{array}{rcl}i\left(t\right)& =& \frac{359.3\cos \left(\omega t\right)}{10\Omega }\\ & =& 35.93\cos \left(\omega t\right)\text{A}\\ & & \end{array}$

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

Therefore, the instantaneous power that is absorbed by the resistor is 6455 + 6455 cos (2ωt) W .

Determine the expression for the instantaneous power absorbed by the capacitor.

(b)

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

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

Solve further as,

p (t) = -2582(2ωt ) W

Therefore, the instantaneous power that is absorbed by the capacitor is

-2582sin (2ωt ) W.

Determine the real power that is absorbed by the capacitor.

(c)

Substitute the values in the equation (3) to solve for the real power absorbed by the resistor.

Therefore, the real power absorbed by the resistor is 6455 W .

Determine the real power that is delivered by the capacitor.

(d)

Substitute the values in the equation (4) to solve for the real power absorbed by the resistor.

Therefore, the real power delivered by the capacitor is 258 VAR .