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 to the voltage.

To find the total impedance \( Z_{total} = Z_{1} + Z_{2} = (3 + j6) + (4 - j3) = (3 + 4) + (j6 - j3) = 7 + j3 \, \Omega \). The magnitude of the total impedance is given by \( |Z_{total}| = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58} \approx 7.62 \, \Omega \). Next, the current \( I \) can be calculated using Ohm's Law, \( I = \frac{V}{Z_{total}} \). The supply voltage is \( V = 120 \, V \), so \( I = \frac{120}{Z_{total}} \). To find \( I \) in its polar form, first we calculate the phase angle of \( Z_{total} \), \( \theta = \tan^{-1} \left(\frac{3}{7}\right) \approx 23.2^\circ \). Therefore, the current will have a phase angle of \( -\theta \). So, the magnitude of the current is \( |I| = \frac{120}{7.62} \approx 15.75 \, A \), and the phase angle relative to the voltage is approximately \( -23.2^\circ \).