A circuit contains a \(100 .-\Omega\) resistor, a \(0.0500-\mathrm{H}\) inductor, a \(0.400-\mu \mathrm{F}\) capacitor, and a source of time-varying emf connected in series. The time-varying emf corresponds to \(V_{\mathrm{rms}}=50.0 \mathrm{~V}\) at a frequency of \(2000 . \mathrm{Hz}\). a) Determine the current in the circuit. b) Determine the voltage drop across each component of the circuit. c) How much power is drawn from the source of emf?

We will start by calculating the impedance of each circuit component: For the resistor (R), the impedance is simply the resistance value itself: \(Z_R = 100\Omega\). For the inductor (L), the impedance is given by the formula: \(Z_L = j\omega L = j(2\pi f)L\), where \(\omega\) is the angular frequency, \(f\) is frequency, and \(j\) is the imaginary unit. Given the frequency \(f = 2000 Hz\) and inductance \(L = 0.0500 H\), we have: \(Z_L = j(2\pi(2000))(0.0500) = j(100\pi)\). For the capacitor (C), the impedance is given by the formula: \(Z_C = -j\frac{1}{\omega C} = -j\frac{1}{2\pi fC}\). Given the frequency \(f = 2000 Hz\) and capacitance \(C = 0.400 \times 10^{-6} F\), we have: \(Z_C = -j\frac{1}{2\pi(2000)(0.4\times10^{-6})} = -j(250\pi)\).

We will now determine the total impedance of the circuit. As all components are connected in series, the total impedance is the sum of individual impedances: \(Z_{total} = Z_R + Z_L + Z_C = 100 + j(100\pi) - j(250\pi)\). \(Z_{total} = 100 - j(150\pi)\)

Now, we calculate the current flowing through the circuit using Ohm's Law: \(I = \frac{V}{Z} = \frac{V_{rms}}{|Z_{total}|}\) First, we need to find the magnitude of the total impedance \(|Z_{total}| = \sqrt{(100)^2 + (-150\pi)^2} \approx 155.70\Omega\). Given the RMS voltage \(V_{rms} = 50 V\), we have: \(I = \frac{50}{155.70} \approx 0.321 A\)

Now, we calculate the individual voltage drops across each component using Ohm's Law: \(V_i = I \times Z_i\) Voltage drop across the resistor: \(V_R = I \times Z_R = (0.321)(100) \approx 32.1 V\) Voltage drop across the inductor: \(V_L = I \times Z_L = (0.321)(j100\pi) \approx j(100.48) V\) Voltage drop across the capacitor: \(V_C = I \times Z_C = (0.321)(-j250\pi) \approx -j(251.20) V\)

Finally, we calculate the power drawn from the source using the formula: \(P = V_{rms} \times I_{rms} \times \cos{\phi}\) where \(\phi\) is the phase difference between the current and voltage. The current phase is given by the angle of the total impedance, which we can express as \(\phi = \arctan{(\frac{-150\pi}{100})} \approx -56.31^{\circ}\). The EMF is considered the reference, so its phase is zero. Therefore, the cosine of the phase difference between the EMF and current is \(\cos{(-56.31^{\circ})} \approx 0.555\). The power drawn from the source of emf is: \(P = (50)(0.321)(0.555) \approx 8.88 W\)