\( Z_{2}=(4-j 3) \Omega \) are connected in series to a supply voltage of 120 V. Determine the magnitude of the current and its phase angle relative to the voltage. [15.76 A, 23.20 lagging]

The magnitude of the current in a series circuit can be calculated using Ohm's Law, \( I = \frac{V}{Z} \). In this case, the supply voltage \( V \) is 120 V and the impedance \( Z_2 \) is \( 4 - j3 \) Ω. To find the magnitude of \( Z_2 \), we calculate \( |Z_2| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5 \) Ω. Thus, the current \( I = \frac{120}{5} = 24 \) A. The impedance's phase angle \( \theta \) is \( \tan^{-1}\left(\frac{-3}{4}\right) \), approximately -36.87°. Hence, the phase angle relative to the voltage is lagging. In practical terms, knowing the phase angle helps in understanding how the current behaves compared to the voltage supply. In a purely resistive load, voltage and current are in phase, resulting in maximum power transfer. However, with reactance (like in this case with \( Z_2 \)), the current lags the voltage, affecting both the power factor and system efficiency. This is crucial for electric circuits in industries, as optimal power management reduces costs and improves system reliability!