937_Physics ProblemsTechnical Physics

# 937_Physics ProblemsTechnical Physics - 278 P33.47...

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278 Alternating Current Circuits P33.47 The rms voltage across the transformer primary is N N V 1 2 2, rms di so the source voltage is ∆∆ VI R N N V ss , rms , rms , rms =+ 1 1 2 2 . The secondary current is V R L 2, rms , so the primary current is N N V R I L 2 1 2 1 , rms , rms = . FIG. P33.47 Then V NV R NR N s s L , rms , rms , rms 22 1 12 2 didi and R V N s L s =− F H G G I K J J F H G I K J = 1 2 5500 2250 80 0 5250 2 87 5 , rms , rms rms V V V af , . . . . .. P33.48 (a) V N N V 2 2 1 1 , rms , rms = N N V V 2 1 2 1 3 10 0 10 83 3 == × = , , . . rms rms V 120 V (b) IV 0900 , rms , rms 1, rms 1, rms = . I 2, rms V V 24.0 V 10 0 10 0 900 120 120 3 ×= F H G I K J ej a f I 2, rms mA = 54 0 . (c) Z V I 2 2 2 3 10 0 10 185 × = , rms , rms V 0.054 A k . P33.49 (a) R × = 4 50 10 6 44 10 290 45 M m ΩΩ e j and I V rms rms 5 W 5.00 10 V A × × = P 500 10 10 0
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