Question

Repeat Problem 2.12 if the resistor and capacitor are connected in series.

Short answer

1. The instantaneous power absorbed by the resistor is 889.994 + 889.994 cos (2ωt) W .
2. The instantaneous power absorbed by the capacitor is 2225.14sin (2ωt) W .
3. The real power absorbed by the resistor is 889.8 W .
4. The reactive power absorbed by the capacitoris 224 VAR .
5. The load power factor is 0.37 leading.

Step-by-step solution

Determine the formulas of instantaneous current, instantaneous power, real power, reactive power and power factor.

Write the formula of instantaneous current

.

...............(1)

Write the formula of instantaneous power.

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

Write the formula of real power absorbed by resistor.

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

Write the formula of reactive power absorbed by capacitor.

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

Write the formula of power factor.

……. (5)

Determine the instantaneous power absorbed by the resistor and capacitor.

Determine the instantaneous current.

Substitute 359.3cosωt for v(t) and 10-j25 for R-jX_c in equation (1).

Determine the instantaneous power.

Substitute 359.3cosωt for V(t) and 13.34cos (ωt +68.2^o) for I (t) in equation (2).

Solve further as,

(a) The instantaneous power absorbed by the resistor is 889.994 +889.994cos (2ωt ) W

(b) The instantaneous power absorbed by the capacitor is 2225.14sin (2ωt ) W

Determine thereal power absorbed by the resistor.

(c)

Substitute 13.34A for I_max and 10Ω for R in equation (3).

Determine the reactive power absorbed by the capacitor.

(d)

Substitute 13.34A for I_max and 25Ω for X_c in equation (4).

Determine the load power factor.

(e)

Substitute 889.8 W for P and 2224 VAR for Q in equation (5).