Question

When a resistor is connected by itself to an ac generator, the average power delivered to the resistor is 6.500 W. When a capacitor is added in series with the resistor, the power delivered is 4.875 W. When an inductor is added in series with the resistor (without the capacitor), the power delivered is 2.925 W. Determine the power delivered when both the capacitor and the inductor are added in series with the resistor?

Please solve with steps.

Answer 1

Answer 2

P1= 6.5 =V^2/R

P2=4.785 =( V^2/(R^2+Xc^2))*R

P3=2.925= (V^2/(R^2+Xl^2))*R

P4= (V^2/(R^2+Xl^2+Xc^2))*R

P4= (V^2/(R^2+Xl^2+R^2+Xc^2-R^2))*R

= (V^2/ ([ (V^2*R/P3) + (V^2*R/P2) - V^2/P1] )*R

=1/[(1/P3) + (1/P2) - (1/P1)]

=1.419 W

Answer 3