To solve the problem step by step, we will first analyze the two impedances \( Z_1 \) and \( Z_2 \) separately, calculate their resistances and reactances, and then find the total power absorbed and overall power factor when they are connected in series. Finally, we will determine the reactance needed to raise the overall power factor to unity.
### Step 1: Calculate the impedance \( Z_1 \)
Given:
- Power \( P_1 = 100 \, \text{W} \)
- Voltage \( V = 230 \, \text{V} \)
- Power factor \( \text{pf}_1 = 0.5 \) (lagging)
Using the formula for power:
\[
P = V \cdot I \cdot \text{pf}
\]
We can express the current \( I_1 \):
\[
I_1 = \frac{P_1}{V \cdot \text{pf}_1} = \frac{100}{230 \cdot 0.5} = \frac{100}{115} \approx 0.8696 \, \text{A}
\]
Now, calculate the impedance \( Z_1 \):
\[
Z_1 = \frac{V}{I_1} = \frac{230}{0.8696} \approx 264.5 \, \Omega
\]
### Step 2: Calculate the resistance \( R_1 \) and reactance \( X_L \) for \( Z_1 \)
Using the power factor:
\[
R_1 = Z_1 \cdot \text{pf}_1 = 264.5 \cdot 0.5 = 132.25 \, \Omega
\]
Now, calculate the reactance \( X_L \):
\[
X_L = \sqrt{Z_1^2 - R_1^2} = \sqrt{(264.5)^2 - (132.25)^2} \approx 229 \, \Omega
\]
### Step 3: Calculate the impedance \( Z_2 \)
Given:
- Power \( P_2 = 60 \, \text{W} \)
- Power factor \( \text{pf}_2 = 0.6 \) (leading)
Using the same method:
\[
I_2 = \frac{P_2}{V \cdot \text{pf}_2} = \frac{60}{230 \cdot 0.6} = \frac{60}{138} \approx 0.4348 \, \text{A}
\]
Now, calculate the impedance \( Z_2 \):
\[
Z_2 = \frac{V}{I_2} = \frac{230}{0.4348} \approx 529 \, \Omega
\]
### Step 4: Calculate the resistance \( R_2 \) and reactance \( X_C \) for \( Z_2 \)
Using the power factor:
\[
R_2 = Z_2 \cdot \text{pf}_2 = 529 \cdot 0.6 \approx 317.4 \, \Omega
\]
Now, calculate the reactance \( X_C \):
\[
X_C = \sqrt{Z_2^2 - R_2^2} = \sqrt{(529)^2 - (317.4)^2} \approx 423.2 \, \Omega
\]
### Step 5: Calculate total resistance \( R \) and total reactance \( X \) when \( Z_1 \) and \( Z_2 \) are in series
Total resistance:
\[
R = R_1 + R_2 = 132.25 + 317.4 \approx 449.65 \, \Omega
\]
Total reactance:
\[
X = X_L - X_C = 229 - 423.2 \approx -194.2 \, \Omega
\]
### Step 6: Calculate total impedance \( Z \)
Using the formula:
\[
Z = \sqrt{R^2 + X^2} = \sqrt{(449.65)^2 + (-194.2)^2} \approx 489.79 \, \Omega
\]
### Step 7: Calculate total power absorbed \( P \) and overall power factor \( \text{pf} \)
Total current \( I \):
\[
I = \frac{V}{Z} = \frac{230}{489.79} \approx 0.469 \, \text{A}
\]
Total power:
\[
P = V \cdot I \cdot \text{pf} = 230 \cdot 0.469 \cdot \frac{R}{Z} = 230 \cdot 0.469 \cdot \frac{449.65}{489.79} \approx 99.4 \, \text{W}
\]
Overall power factor:
\[
\text{pf} = \frac{R}{Z} = \frac{449.65}{489.79} \approx 0.92
\]
### Step 8: Calculate the value of reactance to be added for unity power factor
To achieve unity power factor, we need:
\[
X + X_{added} = 0 \implies X_{added} = -X
\]
\[
X_{added} = 194.2 \, \Omega
\]
### Final Answers
(a) Total power absorbed: \( 99.4 \, \text{W} \), Overall power factor: \( 0.92 \)
(b) Reactance to be added: \( 194.2 \, \Omega \)
