Vector Subtraction and Conversion to Polar Form
Two vectors A=20+j30,B= -10-j15.find A-B and express the result in polar form
Problem Statement
Two vectors are given as:
- \(\vec{A} = 20 + j30\)
- \(\vec{B} = -10 - j15\)
Find the value of \(\vec{A} - \vec{B}\) and express the resulting vector in polar form (\(r\angle\theta\)).
Step 1: Perform Vector Subtraction (\(\vec{A} - \vec{B}\))
Subtract the real and imaginary components independently:
$$\vec{A} - \vec{B} = (20 + j30) - (-10 - j15)$$
$$\vec{A} - \vec{B} = [20 - (-10)] + j[30 - (-15)]$$
$$\vec{A} - \vec{B} = (20 + 10) + j(30 + 15)$$
$$\vec{A} - \vec{B} = \mathbf{30 + j45}$$
Step 2: Conversion to Polar Form (\(r\angle\theta\))
For any complex number in rectangular form \(Z = x + jy\), the polar representation is given by:
$$Z = r\angle\theta$$
1. Magnitude (\(r\)):
$$r = \sqrt{x^2 + y^2} = \sqrt{(30)^2 + (45)^2}$$
$$r = \sqrt{900 + 2025} = \sqrt{2925} \approx \mathbf{54.08}$$
2. Phase Angle (\(\theta\)):
Since both the real part (\(x = 30 > 0\)) and the imaginary part (\(y = 45 > 0\)) are positive, the vector lies in the first quadrant:
$$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{45}{30}\right) = \tan^{-1}(1.5)$$
$$\theta \approx \mathbf{56.31^\circ} \quad (\text{or } 0.9828\text{ radians})$$
Phasor Representation of the Resultant
- Rectangular Form: \(\vec{A} - \vec{B} = \mathbf{30 + j45}\)
- Polar Form: \(\vec{A} - \vec{B} = \mathbf{54.08 \angle 56.31^\circ}\)
- Exponential Form: \(\vec{A} - \vec{B} = 54.08 \, e^{j 0.9828}\)
Complex Number Operations: Magnitude and Slope
Perform the following operations and find magnitude and slope in
each case.(a)A+B(b)A-B (c) AB, where A=20+j15 and
B= 30-j4
Given Data
- \(A = 20 + j15\)
- \(B = 30 - j4\)
For any complex number \(Z = x + jy\):
- Magnitude: \(\vert{}Z\vert{} = \sqrt{x^2 + y^2}\)
- Slope (\(m\)): \(m = \tan\theta = \frac{y}{x} = \frac{\text{Imaginary part}}{\text{Real part}}\)
- Phase Angle (\(\theta\)): \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)
(a) Operation: \(A + B\)
1. Addition in Rectangular Form:
$$A + B = (20 + j15) + (30 - j4)$$
$$A + B = (20 + 30) + j(15 - 4) = \mathbf{50 + j11}$$
2. Magnitude:
$$\vert{}A + B\vert{} = \sqrt{(50)^2 + (11)^2} = \sqrt{2500 + 121} = \sqrt{2621} \approx \mathbf{51.20}$$
3. Slope:
$$\text{Slope } (m) = \frac{y}{x} = \frac{11}{50} = \mathbf{0.22}$$
Phase Angle: \(\theta = \tan^{-1}(0.22) \approx 12.41^\circ\)
(b) Operation: \(A - B\)
1. Subtraction in Rectangular Form:
$$A - B = (20 + j15) - (30 - j4)$$
$$A - B = (20 - 30) + j(15 - (-4)) = \mathbf{-10 + j19}$$
2. Magnitude:
$$\vert{}A - B\vert{} = \sqrt{(-10)^2 + (19)^2} = \sqrt{100 + 361} = \sqrt{461} \approx \mathbf{21.47}$$
3. Slope:
$$\text{Slope } (m) = \frac{y}{x} = \frac{19}{-10} = \mathbf{-1.90}$$
Phase Angle (Second Quadrant): \(\theta = 180^\circ - \tan^{-1}\left(\frac{19}{10}\right) = 180^\circ - 62.24^\circ \approx 117.76^\circ\)
(c) Operation: \(A \times B\)
1. Multiplication in Rectangular Form:
$$A \cdot B = (20 + j15)(30 - j4)$$
$$A \cdot B = (20 \times 30) - j(20 \times 4) + j(15 \times 30) - j^2(15 \times 4)$$
Since \(j^2 = -1\):
$$A \cdot B = 600 - j80 + j450 - (-1)(60)$$
$$A \cdot B = (600 + 60) + j(450 - 80) = \mathbf{660 + j370}$$
2. Magnitude:
$$\vert{}A \cdot B\vert{} = \sqrt{(660)^2 + (370)^2} = \sqrt{435600 + 136900} = \sqrt{572500} \approx \mathbf{756.64}$$
(Verification: \(|A| \times |B| = \sqrt{20^2 + 15^2} \times \sqrt{30^2 + (-4)^2} = 25 \times \sqrt{
3-Phase Balanced Star Connected Load Calculation
Problem Statement
A balanced star-connected load of \((8 + j6)\ \Omega\) per phase is connected to a \(3\)-phase, \(230\text{ V}\) supply.
Find:
- Line current (\(I_L\))
- Active power (\(P\))
- Reactive power (\(Q\))
Given Data
- Line Voltage (\(V_L\)) = \(230\text{ V}\)
- Impedance per phase (\(\vec{Z}_{ph}\)) = \((8 + j6)\ \Omega\)
- Resistance per phase (\(R_{ph}\)) = \(8\ \Omega\)
- Inductive Reactance per phase (\(X_{ph}\)) = \(6\ \Omega\)
- Connection type = Star (\(\text{Y}\)) connection
Step 1: Phase Impedance and Power Factors
Magnitude of phase impedance (\(Z_{ph}\)):
$$Z_{ph} = \sqrt{R_{ph}^2 + X_{ph}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10\ \Omega}$$
Power factor of the circuit (\(\cos\phi\)):
$$\cos\phi = \frac{R_{ph}}{Z_{ph}} = \frac{8}{10} = \mathbf{0.8} \quad (\text{lagging})$$
Reactive factor of the circuit (\(\sin\phi\)):
$$\sin\phi = \frac{X_{ph}}{Z_{ph}} = \frac{6}{10} = \mathbf{0.6}$$
Step 2: Line Current (\(I_L\))
In a star connection, the phase voltage (\(V_{ph}\)) is given by:
$$V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{230}{\sqrt{3}} \approx \mathbf{132.79\text{ V}}$$
Phase current (\(I_{ph}\)):
$$I_{ph} = \frac{V_{ph}}{Z_{ph}} = \frac{132.79}{10} = \mathbf{13.28\text{ A}}$$
For a star connection, the line current equals the phase current (\(I_L = I_{ph}\)):
$$I_L = I_{ph} = \mathbf{13.28\text{ A}}$$
Step 3: Active Power (\(P\))
The total active (true) power drawn by the balanced 3-phase load is:
$$P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\phi$$
$$P = \sqrt{3} \times 230 \times 13.28 \times 0.8 \approx \mathbf{4232\text{ W}} \quad (\text{or } 4.232\text{ kW})$$
(Alternate check: \(P = 3 \cdot I_{ph}^2 \cdot R_{ph} = 3 \times (13.28)^2 \times 8 \approx 4232\text{ W}\))
Step 4: Reactive Power (\(Q\))
The total reactive power drawn by the balanced 3-phase load is:
$$Q = \sqrt{3} \cdot V_L \cdot I_L \cdot \sin\phi$$
$$Q = \sqrt{3} \times 230 \times 13.28 \times 0.6 \approx \mathbf{3174\text{ VAR}} \quad (\text{or } 3.174\text{ kVAR})$$
(Alternate check: \(Q = 3 \cdot I_{ph}^2 \cdot X_{ph} = 3 \times (13.28)^2 \times 6 \approx 3174\text{ VAR}\))
- Line Current (\(I_L\)): \(13.28\text{ A}\)
- Active Power (\(P\)): \(4232\text{ W}\) (or \(4.232\text{ kW}\))
- Reactive Power (\(Q\)): \(3174\text{ VAR}\) (or \(3.174\text{ kVAR}\))
Parallel Circuit Analysis Using Admittance Method
Problem Statement
Two impedances \(Z_1 = (10 + j15)\ \Omega\) and \(Z_2 = (6 - j8)\ \Omega\) are connected in parallel. If the total current supplied is \(15\text{ A}\), determine the current and power taken by each branch using the admittance method.
Given Data
- Branch 1 Impedance: \(\vec{Z}_1 = 10 + j15\ \Omega\) \((R_1 = 10\ \Omega, X_{L1} = 15\ \Omega)\)
- Branch 2 Impedance: \(\vec{Z}_2 = 6 - j8\ \Omega\) \((R_2 = 6\ \Omega, X_{C2} = 8\ \Omega)\)
- Total current magnitude: \(I = 15\text{ A}\)
Step 1: Calculate Admittance of Each Branch
A. Admittance of Branch 1 (\(\vec{Y}_1\))
$$\vec{Y}_1 = \frac{1}{\vec{Z}_1} = \frac{1}{10 + j15}$$
Rationalizing the denominator:
$$\vec{Y}_1 = \frac{10 - j15}{(10 + j15)(10 - j15)} = \frac{10 - j15}{10^2 + 15^2} = \frac{10 - j15}{100 + 225} = \frac{10 - j15}{325}$$
$$\vec{Y}_1 = 0.0308 - j0.0462\ \mho\ (\text{S})$$
Magnitude of \(\vec{Y}_1\):
$$|\vec{Y}_1| = \sqrt{(0.0308)^2 + (-0.0462)^2} = \frac{1}{|\vec{Z}_1|} = \frac{1}{\sqrt{325}} \approx \mathbf{0.0555\ \mho}$$
B. Admittance of Branch 2 (\(\vec{Y}_2\))
$$\vec{Y}_2 = \frac{1}{\vec{Z}_2} = \frac{1}{6 - j8}$$
Rationalizing the denominator:
$$\vec{Y}_2 = \frac{6 + j8}{(6 - j8)(6 + j8)} = \frac{6 + j8}{6^2 + 8^2} = \frac{6 + j8}{36 + 64} = \frac{6 + j8}{100}$$
$$\vec{Y}_2 = 0.0600 + j0.0800\ \mho\ (\text{S})$$
Magnitude of \(\vec{Y}_2\):
$$|\vec{Y}_2| = \sqrt{(0.06)^2 + (0.08)^2} = \frac{1}{|\vec{Z}_2|} = \frac{1}{10} = \mathbf{0.1000\ \mho}$$
Step 2: Calculate Total Admittance (\(\vec{Y}_T\))
The total equivalent admittance of the parallel combination is the sum of the individual branch admittances:
$$\vec{Y}_T = \vec{Y}_1 + \vec{Y}_2 = (0.0308 - j0.0462) + (0.0600 + j0.0800)$$
$$\vec{Y}_T = (0.0308 + 0.0600) + j(-0.0462 + 0.0800) = \mathbf{0.0908 + j0.0338\ \mho}$$
Magnitude of total admittance (\(|\vec{Y}_T|\)):
$$|\vec{Y}_T| = \sqrt{(0.0908)^2 + (0.0338)^2} = \sqrt{0.008245 + 0.001142} = \sqrt{0.009387} \approx \mathbf{0.0969\ \mho}$$
Step 3: Calculate Supply Voltage (\(V\))
Since total current is \(I = V \cdot |\vec{Y}_T|\):
$$V = \frac{I}{|\vec{Y}_T|} = \frac{15}{0.0969} \approx \mathbf{154.80\text{ V}}$$
Step 4: Current in Each Branch
-
Current in Branch 1 (\(I_1\)):
$$I_1 = V \cdot |\vec{Y}_1| = 154.80 \times 0.0555 \approx \mathbf{8.59\text{ A}}$$
-
Current in Branch 2 (\(I_2\)):
$$I_2 = V \cdot |\vec{Y}_2| = 154.80 \times 0.1000 \approx \mathbf{15.48\text{ A}}$$
Step 5: Power Taken by Each Branch
Active power consumed in any AC branch is given by \(P = I^2 R\):
-
Power taken by Branch 1 (\(P_1\)):
$$P_1 = I_1^2 \cdot R_1 = (8.59)^2 \times 10 = 73.788 \times 10 \approx \mathbf{737.9\text{ W}}$$
-
Power taken by Branch 2 (\(P_2\)):
$$P_2 = I_2^2 \cdot R_2 = (15.48)^2 \times 6 = 239.63 \times 6 \approx \mathbf{1437.8\text{ W}}$$
- Branch 1 Current (\(I_1\)): \(8.59\text{ A}\)
- Branch 2 Current (\(I_2\)): \(15.48\text{ A}\)
- Branch 1 Power (\(P_1\)): \(737.9\text{ W}\)
- Branch 2 Power (\(P_2\)): \(1437.8\text{ W}\)
- (Total Power Consumed, \(P_T = P_1 + P_2 \approx 2175.7\text{ W}\))