Question

An AC source has a frequency f = 1200 HZ and peak voltage 3.5 V. It is connected in series to a resistor resistance R = 15 Ohms, an inductor of inductance L = 5 mH and a capacitor of capacitance C = 2 microfarads.

How would you optimize the power factor ( make it equal to 1)?

a. Add a  capacitor in series with the existing one

b. Add a  capacitor in parallel with the existing one

c. Add a  capacitor in parallel with the existing one

d. Add a  capacitor in series with the existing one

~~The correct answer is D but when i solved for X_L=X_C I found that it should have been B. Why is the answer D?

Answer 1

We know that for power factor to be 1, X_​​l​​​​ = X_{​​​​​​c}

the capacitor we are having is 2 micro farad and the capacitance we need is 3.51 micro farad which is bigger than what we have.

Capacitors if connected in series the resultant will be smaller than the smallest .So, there is no way getting a bigger capacitance by connecting in series .

We can get bigger capacitance by connecting another capacitor in parellel