Question Three Two impedances, \( Z_{1}=(3+j 6) \Omega \) and \( Z_{2}=(4-j 3) \Omega \) are connected in series to a supply voltage of 120 V . Determine the magnitude of the current and phase angle relative the voltage.

To find the total impedance \( Z_{total} = Z_{1} + Z_{2} = (3+j6) + (4-j3) = (7+j3) \Omega \). Next, we calculate the magnitude of the total impedance: \[ |Z_{total}| = \sqrt{(7)^2 + (3)^2} = \sqrt{49 + 9} = \sqrt{58} \approx 7.62 \, \Omega. \] Now, use Ohm's law to find the current: \[ I = \frac{V}{Z_{total}} = \frac{120 V}{7.62 \,\Omega} \approx 15.74 \, A. \] The phase angle \( \phi \) can be calculated using: \[ \phi = \tan^{-1} \left( \frac{\text{Im}(Z_{total})}{\text{Re}(Z_{total})} \right) = \tan^{-1} \left( \frac{3}{7} \right) \approx 23.2°. \] So, the magnitude of the current is approximately \( 15.74 \, A \) and the phase angle relative to the voltage is about \( 23.2° \). Now go ahead and plug in the numbers for a practical understanding! Exploring these calculations can give you insights into how electrical loads operate in real circuits, making your next electrical project a breeze. Similarly, realizing how to correctly combine impedances can save you time and troubleshooting down the line—who wants to redo calculations over and over? Keep practicing!