Question: 5. Find IL and the circuit pf.
120 V
R = 45 \Omega
XL = 105 \Omega
top-left: 120 V source feeding a parallel circuit with a 45 \Omega resistor in the center branch and an inductor with XL = 105 \Omega in the right branch.
6. Find IL and the circuit pf.
120 V
R = 45 \Omega
R = 10 \Omega
XL = 50 \Omega
bottom-left: 120 V source feeding two parallel branches, one branch a 45 \Omega resistor and the other branch a series R-L branch with R = 10 \Omega and XL = 50 \Omega.
1. **Problem Statement:**
Find the current through the inductor $I_L$ and the power factor (pf) of the circuit for problem 5.
2. **Given Data:**
- Voltage source $V = 120$ V
- Resistor $R = 45 \Omega$
- Inductive reactance $X_L = 105 \Omega$
3. **Circuit Description:**
The circuit has two parallel branches:
- Branch 1: Resistor $R = 45 \Omega$
- Branch 2: Inductor with reactance $X_L = 105 \Omega$
4. **Formulas and Rules:**
- Impedance of resistor branch: $Z_R = R = 45 \Omega$
- Impedance of inductor branch: $Z_L = jX_L = j105 \Omega$
- Total admittance $Y_{total} = Y_R + Y_L = \frac{1}{Z_R} + \frac{1}{Z_L}$
- Total current $I_{total} = \frac{V}{Z_{total}}$ where $Z_{total} = \frac{1}{Y_{total}}$
- Current through inductor branch $I_L = V / Z_L$
- Power factor $pf = \cos(\theta)$ where $\theta$ is the phase angle of total impedance
5. **Calculate admittances:**
$$
Y_R = \frac{1}{45} = 0.02222\,S
$$
$$
Y_L = \frac{1}{j105} = -j\frac{1}{105} = -j0.009524\,S
$$
6. **Total admittance:**
$$
Y_{total} = 0.02222 - j0.009524\,S
$$
7. **Calculate total impedance:**
$$
Z_{total} = \frac{1}{Y_{total}} = \frac{1}{0.02222 - j0.009524}
$$
Multiply numerator and denominator by the complex conjugate:
$$
Z_{total} = \frac{0.02222 + j0.009524}{(0.02222)^2 + (0.009524)^2} = \frac{0.02222 + j0.009524}{0.0004933 + 0.0000907} = \frac{0.02222 + j0.009524}{0.000584}
$$
$$
Z_{total} = 38.05 + j16.31 \Omega
$$
8. **Calculate magnitude and phase angle of $Z_{total}$:**
$$
|Z_{total}| = \sqrt{38.05^2 + 16.31^2} = \sqrt{1447.8 + 266.0} = \sqrt{1713.8} = 41.41 \Omega
$$
$$
\theta = \tan^{-1}\left(\frac{16.31}{38.05}\right) = \tan^{-1}(0.4287) = 23.2^\circ
$$
9. **Calculate total current:**
$$
I_{total} = \frac{V}{|Z_{total}|} = \frac{120}{41.41} = 2.90\,A
$$
10. **Calculate power factor:**
$$
pf = \cos(23.2^\circ) = 0.92 \text{ (lagging)}
$$
11. **Calculate current through inductor branch:**
$$
I_L = \frac{V}{|Z_L|} = \frac{120}{105} = 1.14\,A
$$
The current through the inductor leads the voltage by $90^\circ$ because it is purely inductive.
**Final answers:**
- $I_L = 1.14$ A
- Power factor $pf = 0.92$ lagging