Answer :
Solution:
Given:
- Resistor ([tex]\( R \)[/tex]) = 2002 ohms
- Inductance ([tex]\( L \)[/tex]) = 2.387 mH = 2.387 × 10^-3 H
- Supply Voltage ([tex]\( V \)[/tex]) = 60 V
- Supply Frequency ([tex]\( f \)[/tex]) = 1 kHz = 1000 Hz
### Step-by-Step Solution:
(a) The current in each branch:
1. Current through the resistor branch:
Using Ohm's Law:
[tex]\[ I_R = \frac{V}{R} \][/tex]
Plug in the values:
[tex]\[ I_R = \frac{60 \, \text{V}}{2002 \, \Omega} \approx 0.02997 \, \text{A} \][/tex]
2. Current through the inductor branch:
First, compute the inductive reactance ([tex]\(X_L\)[/tex]):
[tex]\[ X_L = 2 \pi f L \][/tex]
Plug in the values:
[tex]\[ X_L = 2 \pi \times 1000 \, \text{Hz} \times 2.387 \times 10^{-3} \, \text{H} \approx 14.997 \, \Omega \][/tex]
Using Ohm's Law:
[tex]\[ I_L = \frac{V}{X_L} \][/tex]
Plug in the values:
[tex]\[ I_L = \frac{60 \, \text{V}}{14.997 \, \Omega} \approx 4.00054 \, \text{A} \][/tex]
(b) The supply current:
Since the resistor and inductor are in parallel, the supply current ([tex]\(I_{s}\)[/tex]) is the vector sum of the currents in each branch. For the magnitude:
[tex]\[ I_{s} = \sqrt{I_{R}^2 + I_{L}^2} \][/tex]
Substitute the values:
[tex]\[ I_{s} = \sqrt{(0.02997 \, \text{A})^2 + (4.00054 \, \text{A})^2} \approx 4.00066 \, \text{A} \][/tex]
(c) The circuit phase angle:
The phase angle ([tex]\(\theta\)[/tex]) between the supply current and the supply voltage is given by:
[tex]\[ \tan(\theta) = \frac{I_L}{I_R} \][/tex]
Substitute the values:
[tex]\[ \tan(\theta) = \frac{4.00054}{0.02997} \][/tex]
Thus, the phase angle is:
[tex]\[ \theta = \arctan\left(\frac{4.00054}{0.02997}\right) \approx 1.5633 \, \text{radians} \][/tex]
(d) The circuit impedance:
The total impedance ([tex]\(Z\)[/tex]) of the circuit can be calculated using:
[tex]\[ Z = \frac{V}{I_s} \][/tex]
Substitute the values:
[tex]\[ Z = \frac{60 \, \text{V}}{4.00066 \, \text{A}} \approx 14.9975 \, \Omega \][/tex]
(e) The power consumed:
The power consumed ([tex]\(P\)[/tex]) in the circuit (which is the real power) is:
[tex]\[ P = V \cdot I_R \cdot \cos(\theta) \][/tex]
We already have the power factor ([tex]\(\cos(\theta)\)[/tex]):
[tex]\[ \cos(\theta) = \frac{I_R}{I_s} \][/tex]
Let's calculate:
[tex]\[ \cos(\theta) = \frac{0.02997 \, \text{A}}{4.00066 \, \text{A}} \approx 0.0075 \][/tex]
Now substitute into the power formula:
[tex]\[ P = (60 \, \text{V}) \times (0.02997 \, \text{A}) \times 0.0075 \approx 0.01347 \, \text{W} \][/tex]
### Summary of Results:
(a) Current in the resistor branch: [tex]\(I_R \approx 0.02997 \, \text{A}\)[/tex]
Current in the inductor branch: [tex]\(I_L \approx 4.00054 \, \text{A}\)[/tex]
(b) Supply current: [tex]\(I_s \approx 4.00066 \, \text{A}\)[/tex]
(c) Circuit phase angle: [tex]\(\theta \approx 1.5633 \, \text{radians}\)[/tex]
(d) Circuit impedance: [tex]\(Z \approx 14.9975 \, \Omega\)[/tex]
(e) Power consumed: [tex]\(P \approx 0.01347 \, \text{W}\)[/tex]
Given:
- Resistor ([tex]\( R \)[/tex]) = 2002 ohms
- Inductance ([tex]\( L \)[/tex]) = 2.387 mH = 2.387 × 10^-3 H
- Supply Voltage ([tex]\( V \)[/tex]) = 60 V
- Supply Frequency ([tex]\( f \)[/tex]) = 1 kHz = 1000 Hz
### Step-by-Step Solution:
(a) The current in each branch:
1. Current through the resistor branch:
Using Ohm's Law:
[tex]\[ I_R = \frac{V}{R} \][/tex]
Plug in the values:
[tex]\[ I_R = \frac{60 \, \text{V}}{2002 \, \Omega} \approx 0.02997 \, \text{A} \][/tex]
2. Current through the inductor branch:
First, compute the inductive reactance ([tex]\(X_L\)[/tex]):
[tex]\[ X_L = 2 \pi f L \][/tex]
Plug in the values:
[tex]\[ X_L = 2 \pi \times 1000 \, \text{Hz} \times 2.387 \times 10^{-3} \, \text{H} \approx 14.997 \, \Omega \][/tex]
Using Ohm's Law:
[tex]\[ I_L = \frac{V}{X_L} \][/tex]
Plug in the values:
[tex]\[ I_L = \frac{60 \, \text{V}}{14.997 \, \Omega} \approx 4.00054 \, \text{A} \][/tex]
(b) The supply current:
Since the resistor and inductor are in parallel, the supply current ([tex]\(I_{s}\)[/tex]) is the vector sum of the currents in each branch. For the magnitude:
[tex]\[ I_{s} = \sqrt{I_{R}^2 + I_{L}^2} \][/tex]
Substitute the values:
[tex]\[ I_{s} = \sqrt{(0.02997 \, \text{A})^2 + (4.00054 \, \text{A})^2} \approx 4.00066 \, \text{A} \][/tex]
(c) The circuit phase angle:
The phase angle ([tex]\(\theta\)[/tex]) between the supply current and the supply voltage is given by:
[tex]\[ \tan(\theta) = \frac{I_L}{I_R} \][/tex]
Substitute the values:
[tex]\[ \tan(\theta) = \frac{4.00054}{0.02997} \][/tex]
Thus, the phase angle is:
[tex]\[ \theta = \arctan\left(\frac{4.00054}{0.02997}\right) \approx 1.5633 \, \text{radians} \][/tex]
(d) The circuit impedance:
The total impedance ([tex]\(Z\)[/tex]) of the circuit can be calculated using:
[tex]\[ Z = \frac{V}{I_s} \][/tex]
Substitute the values:
[tex]\[ Z = \frac{60 \, \text{V}}{4.00066 \, \text{A}} \approx 14.9975 \, \Omega \][/tex]
(e) The power consumed:
The power consumed ([tex]\(P\)[/tex]) in the circuit (which is the real power) is:
[tex]\[ P = V \cdot I_R \cdot \cos(\theta) \][/tex]
We already have the power factor ([tex]\(\cos(\theta)\)[/tex]):
[tex]\[ \cos(\theta) = \frac{I_R}{I_s} \][/tex]
Let's calculate:
[tex]\[ \cos(\theta) = \frac{0.02997 \, \text{A}}{4.00066 \, \text{A}} \approx 0.0075 \][/tex]
Now substitute into the power formula:
[tex]\[ P = (60 \, \text{V}) \times (0.02997 \, \text{A}) \times 0.0075 \approx 0.01347 \, \text{W} \][/tex]
### Summary of Results:
(a) Current in the resistor branch: [tex]\(I_R \approx 0.02997 \, \text{A}\)[/tex]
Current in the inductor branch: [tex]\(I_L \approx 4.00054 \, \text{A}\)[/tex]
(b) Supply current: [tex]\(I_s \approx 4.00066 \, \text{A}\)[/tex]
(c) Circuit phase angle: [tex]\(\theta \approx 1.5633 \, \text{radians}\)[/tex]
(d) Circuit impedance: [tex]\(Z \approx 14.9975 \, \Omega\)[/tex]
(e) Power consumed: [tex]\(P \approx 0.01347 \, \text{W}\)[/tex]
