Question

An inductor (L = 365 mH), a capacitor (C = 4.43 uF), and a resistor (R = 6052) are connected in series. A 50.0 Hz AC source produces a peak current of 250 mA in the circuit. (a) Calculate the required peak voltage AV max (b) Determine the phase angle by which the current leads or lags the applied voltage. Step 1 The total impedance depends on the frequency and the resistance of the circuit. The voltage amplitude is in direct proportion to the current and the resistance. The phase angle is between -90° and +90°. Δες Δ, Δει- 2004 R L XLR Х Step 2 We will find the reactances of the inductor and the capacitor and assign these impedance components to the phase directions of the voltages in the phasor diagram. We will combine the impedance components, along with the resistance, by vector addition to find the constant of proportionality between the current and voltage. The direction of the impedance in phase space is the angle between the current and the voltage.

Answer 1

l365 mi с 4.43 ME 6os 2 so H2 . I peak 2 som mA Inductive reactance ,XL = atife = 2lly 50 H2 x 365x108H. 114.612 Capacitive reacdance, Xca atute xo: 2 Rx50x 6.438,00f 2 718.92 x2 I peak xa DUpeak 2: VREM (re-A2DP - Voosit(-609.3)2 855 213.75" supraka asoxiolyess 6) lan q Verve Xc - Xc VR R lan xc-Xc R dan' (-1) - 45 phase angle, q

a sen'es RC when XL L Xciphash angled is nigatie and circuit acts as circuit in which current leads voltage R > q 1xc-XL 3