Question

(t) 185 V sin [27(7.25 MHz)t). Use this information For the RLC circuit at the right R-13612, C-230 JF, L -0.55 p at the right R-136 22, C-230 uF. L-0.55 pH, and to answer problems 9, 10, 11 and 12. [6] 9. Determine the reactance of the capacitor in the circuit Answer: @) 75.7 2. (6) 87.8 2. (c) 95.41. (d) 125 2. (e). 150 u12 () (d) 58. 1 2 [6] 10. Determine the reactance of the inductor in the circuit Answer: (a) 15.1 12. (b) 25.12. (c) 35.1 2. (e) 45. 12. [6] 11. The impedance of the circuit is Answer: (a) 153 12. (b) 171 . () 1842. () 1980 (1) - (d) 2100. () [4] 12. Comparing an electrical RLC circuit and a Mass-Spring system that can move along a surface, (a) Capacitance corresponds to the mass, the location of the mass, the spring constant of the spring). (6) Inductance corresponds to the mass, the location of the mass, the spring constant of the spring). (C) Electrical current corresponds to (position, speed, acceleration of the mass (d) Electric charge corresponds to (position, speed, acceleration) of the mass. [12] 13. A piece of wire is bent in the shape shown (two 650 cm horizontal section and a semicircle section with radius 485 cm). Determine the magnetic field (magnitude and direction) at point P, the center of the semicircle. The current i = 5 mA. (Hint you need Biot-Savart Law.) If you show your work in an organized and clear manner that I can clearly see that you are working the problem correctly and not just writing from the memory of some problem that was previously worked out, you will get partial credit, depending on how correct you work is.

Answer 1

9)

compare  

ε(t) = Vo*sin(ω*t)

so

Xc = 1/Cω = 1/ (230*10^-6 *2*pi*7.25*10^6) =9.54452432e-5 =95.44 μm

10) X_L = L*w = 0.55*10^-12*2*pi*7.25*10^6 =2.50542014e-5 = 25.1 μm

11)

Z = sqrt( 13.6^2+ (9.54452432-2.50542014)^2) = 15.313*10^-5 = 153.13 μm

12)

Electrical Quantity Mechanical Analog (Force-Current) Mechanical Analog II (Force Voltage) Force, f Voltage, e Velocity, v Current, i Force, f Velocity, v Resistance, R Lubricity, 1/B (Inverse friction) Friction, B Capacitance, Mass, M Compliance, 1/K (Inverse spring constant) Inductance, L Mass, M Compliance, 1/K (Inverse spring constant) Lever, L1 L2 Transformer, N1 N2 Lever, L1 L2

13)

What is the magnetic field at the center of an arc of current of radius R? The arc curves through an angle . If the current is counter-clockwise the magnetic field at the center is out of the page. The magnetic field dB from a representative piece of the wire is: dB = Mo I dl * Î 4112 I=R, and dl x Î = di  Our integral for the net field is easy: Bere n Sak В ць Is 41 R2 K where s is the arc length. equal to Re. 410 Therefore B = 41 Rk

field due to two segment will be zero cause   center of circle lies in their line of current  

theta for half circle = pi

put pi in above derivation

B = μ * I / 4R

I = 0.5*10^-3

R = 4.85

μ -= 4pi*10^-7

now plug the value

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