To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.
### Part (a): Determine the Circuit Element to Raise Power Factor
1. **Understanding Power Factor**: The power factor (PF) is defined as the cosine of the phase angle (φ) between the voltage and current in an AC circuit. It can also be expressed as:
\[
\text{PF} = \cos(\phi) = \frac{R}{Z}
\]
where \( R \) is the resistance and \( Z \) is the impedance.
2. **Given Values**:
- Impedance \( Z = 60.0 \, \Omega \)
- Power Factor \( \text{PF} = 0.720 \)
3. **Finding Resistance**: We can rearrange the power factor formula to find \( R \):
\[
R = \text{PF} \times Z = 0.720 \times 60.0 = 43.2 \, \Omega
\]
4. **Calculating Reactance**: The impedance in an AC circuit can also be expressed as:
\[
Z = \sqrt{R^2 + (X_L - X_C)^2}
\]
where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance. Since the voltage lags the current, we know that \( X_C > X_L \).
5. **Finding \( X_C - X_L \)**: We can rearrange the impedance formula to find \( X_C - X_L \):
\[
X_C - X_L = \sqrt{Z^2 - R^2} = \sqrt{60^2 - 43.2^2}
\]
\[
= \sqrt{3600 - 1866.24} = \sqrt{1733.76} \approx 41.64 \, \Omega
\]
6. **Conclusion for Part (a)**: To raise the power factor, we need to increase \( X_L \) (inductive reactance) by adding an inductor in series, as this will help to balance the circuit and improve the power factor.
### Part (b): Determine the Size of the Inductor to Achieve Unity Power Factor
1. **Setting Power Factor to Unity**: For a unity power factor, we need \( \text{PF} = 1 \), which means:
\[
R = Z
\]
2. **Finding New Impedance**: Since \( Z = 60.0 \, \Omega \) and \( R = 43.2 \, \Omega \), we need to adjust the reactance \( X_L \) to make the total impedance equal to the resistance:
\[
X_C - X_L = 0 \quad \text{(for unity power factor)}
\]
3. **Calculating Required Inductance**: Since we already found \( X_C - X_L = 41.64 \, \Omega \), we need to add an inductor such that:
\[
X_L = X_C - 41.64 \, \Omega
\]
4. **Finding Inductance**: The inductive reactance is given by:
\[
X_L = 2 \pi f L
\]
Rearranging this gives:
\[
L = \frac{X_L}{2 \pi f}
\]
Substituting \( f = 50 \, \text{Hz} \):
\[
L = \frac{41.64}{2 \pi \times 50} \approx \frac{41.64}{314.16} \approx 0.132 \, \text{H} \approx 0.133 \, \text{H}
\]
5. **Conclusion for Part (b)**: To raise the power factor to unity, an inductor with an inductance of approximately \( 0.133 \, \text{H} \) should be added in series with the circuit.
