EC. PRB #4. Name PRB #1 PRB #5 PRB #2 PRB #3 PRB #6 PHY 274 PROBLEM SOLVING WORK
ID: 1295393 • Letter: E
Question
EC. PRB #4. Name PRB #1 PRB #5 PRB #2 PRB #3 PRB #6 PHY 274 PROBLEM SOLVING WORKSHOP XI The current amplitude I A con of inductance 88 versus driving angular fre 4 mH and unknown resistance quency ad for a driven RLC and a 0.94 capacitor are circuit is given in Fig. 31-26. 2 connected in series with an a- The inductance is 200 HH, and 2 temating enf of frequency 930 the emf amplitude is 8.0 V. Hz. If the phase constant be- What are (a) C and (b) R? OLLLL tween the applied voltage and 30 the current is 75. what is the resistapce of the coil7 In Fig. 31-7, setR 200n, CF 70.0 HF, L 230 mH, In Fig. 31-, R 150 n, C 4.70. ALF, and L 250 fd 60.0 Hz, and g 36.0 v. What are (a) z, (b) b, and (c0 mEL The generator provides an eznf with mus valtage 75.0 V and frequency 550 Hz (a) What is the mas current? What is i? (d) Draw a phasor diagram. Z ANGLE thermsvoltage across (b R, (c) C, (d)L,(e) CandL together, and and L together? At what average rate is energy dissipated by (g) R, (h) Cand L? An inductor has a reactance of 100 at A100-wlightbub is plugged into a standard 80 Hz. (a) What is its inductance? (b) What is its reactance at 120-v (TITIS) outlet. Find (a) (b) and C the maximum 160 Hz? Dower.Explanation / Answer
(a)
P=iv
i = p/v = 100/120 = 0.833 A
(b)
Imax = 0.833 x sqrt(2) = 1.18A.
(c)
R = 1202/100 = 144
Pmax = (Imax)2 * R = (1.18)2 x 144 = 200W.
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