To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.
### Part (a): Total Power Absorbed and Overall Power Factor
1. **Calculate Impedance \( Z_1 \) for the first impedance (inductor)**:
- Given: Power \( P_1 = 100 \, W \), Voltage \( V = 230 \, V \), Power Factor \( pf_1 = 0.5 \) (lagging).
- The formula for power is:
\[
P = V \cdot I \cdot pf
\]
- Rearranging gives:
\[
I_1 = \frac{P_1}{V \cdot pf_1} = \frac{100}{230 \cdot 0.5} = \frac{100}{115} \approx 0.8696 \, A
\]
- Now, using Ohm's law:
\[
Z_1 = \frac{V}{I_1} = \frac{230}{0.8696} \approx 264.5 \, \Omega
\]
2. **Calculate Resistance \( R_1 \) and Reactance \( X_L \) for \( Z_1 \)**:
- Using the relationship \( pf = \frac{R}{Z} \):
\[
R_1 = pf_1 \cdot Z_1 = 0.5 \cdot 264.5 \approx 132.25 \, \Omega
\]
- Now calculate \( X_L \):
\[
Z_1^2 = R_1^2 + X_L^2 \implies X_L = \sqrt{Z_1^2 - R_1^2} = \sqrt{(264.5)^2 - (132.25)^2} \approx 229 \, \Omega
\]
3. **Calculate Impedance \( Z_2 \) for the second impedance (capacitor)**:
- Given: Power \( P_2 = 60 \, W \), Power Factor \( pf_2 = 0.6 \) (leading).
- Calculate current \( I_2 \):
\[
I_2 = \frac{P_2}{V \cdot pf_2} = \frac{60}{230 \cdot 0.6} = \frac{60}{138} \approx 0.4348 \, A
\]
- Calculate \( Z_2 \):
\[
Z_2 = \frac{V}{I_2} = \frac{230}{0.4348} \approx 529 \, \Omega
\]
4. **Calculate Resistance \( R_2 \) and Reactance \( X_C \) for \( Z_2 \)**:
- Calculate \( R_2 \):
\[
R_2 = pf_2 \cdot Z_2 = 0.6 \cdot 529 \approx 317.4 \, \Omega
\]
- Now calculate \( X_C \):
\[
Z_2^2 = R_2^2 + X_C^2 \implies X_C = \sqrt{Z_2^2 - R_2^2} = \sqrt{(529)^2 - (317.4)^2} \approx 423.2 \, \Omega
\]
5. **Calculate Total Resistance and Reactance in Series**:
- Total Resistance \( R_{total} = R_1 + R_2 = 132.25 + 317.4 \approx 449.65 \, \Omega \)
- Total Reactance \( X_{total} = X_C - X_L = 423.2 - 229 \approx 194.2 \, \Omega \)
6. **Calculate Total Impedance \( Z_{total} \)**:
- Using:
\[
Z_{total} = \sqrt{R_{total}^2 + X_{total}^2} = \sqrt{(449.65)^2 + (194.2)^2} \approx 489.79 \, \Omega
\]
7. **Calculate Overall Power Factor**:
- Power Factor \( pf_{total} = \frac{R_{total}}{Z_{total}} = \frac{449.65}{489.79} \approx 0.92 \)
8. **Calculate Total Power Absorbed**:
- Total Power \( P_{total} = V \cdot I_{total} \cdot pf_{total} \)
- First, find \( I_{total} \):
\[
I_{total} = \frac{V}{Z_{total}} = \frac{230}{489.79} \approx 0.469 \, A
\]
- Now calculate \( P_{total} \):
\[
P_{total} = 230 \cdot 0.469 \cdot 0.92 \approx 99 \, W
\]
### Part (b): Value of Reactance to Raise Power Factor to Unity
1. **Condition for Unity Power Factor**:
- For unity power factor, \( Z = R \), which means:
\[
R_{total} = \sqrt{R_{total}^2 + (X_C - X_L + X_L')^2}
\]
- Where \( X_L' \) is the new inductive reactance to be added.
2. **Set up the equation**:
- Rearranging gives:
\[
X_C - X_L + X_L' = 0 \implies X_L' = X_L - X_C
\]
- Substitute the values:
\[
X_L' = 194.2 \, \Omega
\]
### Final Answers:
(a) Total Power Absorbed: **99 W**, Overall Power Factor: **0.92**
(b) Value of Reactance to be added: **194.2 Ω**
