Question

When an ac source with a peak potential difference of 42V is connected to a series RLC circuit, the peak current is 4 A. The capacitor is C = 20 μF, the frequency is 60 Hz, and the current leads the potential difference by 30 ∘ degrees. Find the resitance and the inductance.

R = Ω

L = H

Answer 1

R L womt Vm =42v Impedance 2= R + j(XL-Xc) =Rt j (CWL- uc), R²+ (WL- wc)2. 121- Im = peak current - 4 Amp Im = Vm 42 121 ✓ R²+ (w Lwc.)? 42 4221 2 V R² + (wh-toc)2 - R + (WL-toc) R² + (w 2-toch = 72 441 4 0 Im = Vm 24. Q = 30° & a ton (WL- we R -B 30° R the "(WL-tuc) -) WLtouc R WL-touca the 30° R V3

into ean Now put w Lotuc = 2 R² + ( R ) 441 4 R + R? 3 441 4 → 4 R² - 4211 = R² = 44183 46x4 44103 4X4 = 2113 =) R = 4. W = 21 f = 21TX 60 - 120IT f = preanency - Goltz = 204F R=21832 20x10-6 7. from ean 11 120H XL- 1 12017X 20X156 23 445 LOLO 1 120HXL 21+ 21 4. 12011X 20x106 = 137.95 120H XL V v 137.95 120 TL .366 La Aor R = 2113 2113R g.ogn 2 L = 366 H