A series combination of a \(22.0-\mathrm{mH}\) inductor and a \(145.0-\Omega\) resistor are connected across the output terminals of an ac generator with peak voltage \(1.20 \mathrm{kV},\) (a) \(\mathrm{At}\) \(f=1250 \mathrm{Hz},\) what are the voltage amplitudes across the inductor and across the resistor? (b) Do the voltage amplitudes add to give the source voltage (i.e., does \(\left.V_{R}+V_{L}=1.20 \mathrm{kV}\right) ?\) Explain. (c) Draw a phasor diagram to show the addition of the voltages.

Question: In a series RL circuit, the peak source voltage is \(1.20 \mathrm{kV}\), the resistance of the resistor is \(42 \Omega\), and the inductor has an inductance of \(150 \mathrm{mH}\). The frequency of the AC source is \(60 \mathrm{Hz}\). Calculate the voltage amplitudes across the inductor and resistor, and verify that they add up to the source voltage. Draw a phasor diagram to explain the voltage addition. Answer: The voltage amplitude across the inductor is \(V_L = 1048.82 \mathrm{V}\), and across the resistor is \(V_R = 521.94 \mathrm{V}\). Yes, their sum equals the source voltage \((V_R + V_L = 1.20 \mathrm{kV})\). The phasor diagram shows the addition of these voltages as vectors in the complex plane, with \(V_R\) being a horizontal vector and \(V_L\) being a vertical vector, adding up to form a right-angled triangle with the source voltage as the hypotenuse.