b) Three impedances are connected in parallel. \( Z_{1}=2 j+5, Z_{2}=3-j, Z_{3}=4 j \). Find the equivalent admittance \( Y \) where \[ Y=\frac{1}{Z_{1}}+\frac{1}{Z_{2}}+\frac{1}{Z_{3}} \] Express the admittance in both rectangular and polar forms. Rectangular form Polar form

To find the equivalent admittance \( Y \), we first need to compute the individual admittances of each impedance: 1. For \( Z_{1} = 2j + 5 \): \[ Y_{1} = \frac{1}{Z_{1}} = \frac{1}{5 + 2j} \] Multiply numerator and denominator by the conjugate: \[ Y_{1} = \frac{5 - 2j}{(5)^{2} + (2)^{2}} = \frac{5 - 2j}{29} = \frac{5}{29} - \frac{2}{29}j \] 2. For \( Z_{2} = 3 - j \): \[ Y_{2} = \frac{1}{Z_{2}} = \frac{1}{3 - j} \] Multiply numerator and denominator by the conjugate: \[ Y_{2} = \frac{3 + j}{(3)^{2} + (1)^{2}} = \frac{3 + j}{10} = \frac{3}{10} + \frac{1}{10}j \] 3. For \( Z_{3} = 4j \): \[ Y_{3} = \frac{1}{Z_{3}} = \frac{1}{4j} = -\frac{j}{4} \] Now we add the individual admittances: \[ Y = Y_{1} + Y_{2} + Y_{3} = \left( \frac{5}{29} - \frac{2}{29}j \right) + \left( \frac{3}{10} + \frac{1}{10}j \right) + \left( 0 - \frac{1}{4}j \right) \] To combine, we need a common denominator. The least common multiple of \( 29, 10, \) and \( 4 \) is \( 580 \). Converting each term: - \( Y_{1} = \frac{5}{29} - \frac{2}{29}j = \frac{100}{580} - \frac{40}{580}j \) - \( Y_{2} = \frac{3}{10} + \frac{1}{10}j = \frac{174}{580} + \frac{58}{580}j \) - \( Y_{3} = 0 - \frac{1}{4}j = 0 - \frac{145}{580}j \) Now, let's sum the real and imaginary parts: Real part: \[ \frac{100}{580} + \frac{174}{580} = \frac{274}{580} \] Imaginary part: \[ -\frac{40}{580} + \frac{58}{580} - \frac{145}{580} = -\frac{127}{580} \] So the equivalent admittance in rectangular form is: \[ Y = \frac{274}{580} - \frac{127}{580}j \] Now, we can convert this to polar form: 1. Find the magnitude: \[ |Y| = \sqrt{\left( \frac{274}{580} \right)^{2} + \left( -\frac{127}{580} \right)^{2}} = \sqrt{\frac{75076 + 16129}{336400}} = \sqrt{\frac{91105}{336400}} \approx 0.494 \] 2. Find the phase angle: \[ \theta = \tan^{-1} \left( \frac{-127/580}{274/580} \right) = \tan^{-1} \left( -\frac{127}{274} \right) \] Calculating \( \theta \): \[ \theta \approx -0.433 \text{ radians} \approx -24.8^\circ \] So the equivalent admittance in polar form is: \[ Y \approx 0.494 \angle -24.8^\circ \] In conclusion, the equivalent admittance is: - Rectangular form: \( Y \approx \frac{137}{290} - \frac{127}{580}j \) - Polar form: \( Y \approx 0.494 \angle -24.8^\circ \)