Trong ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc v\u1eadt l\u00fd l\u1edbp 12 th\u00ec chuy\u00ean \u0111\u1ec1 v\u1ec1 c\u00f4ng su\u1ea5t c\u1ee7a d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u l\u00e0 kh\u00e1 quan tr\u1ecdng. Sau \u0111\u00e2y c\u00e1c b\u1ea1n h\u00e3y c\u00f9ng tamcaotrithuc.com \u00f4n t\u1eadp l\u1ea1i ki\u1ebfn th\u1ee9c v\u1ec1 c\u00e1c c\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t nh\u00e9.<\/p>\n

I\/ C\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t<\/strong><\/h2>\n

1\/ \u0110\u1ecbnh ngh\u0129a c\u00f4ng su\u1ea5t c\u1ee7a d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u<\/strong><\/h3>\n

c\u00f4ng su\u1ea5t c\u1ee7a 1 d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u ch\u00ednh l\u00e0 c\u00f4ng su\u1ea5t \u0111\u01b0\u1ee3c t\u00ednh trong th\u1eddi gian d\u00e0i hay l\u00e0 c\u00f4ng su\u1ea5t \u0111\u01b0\u1ee3c t\u00ednh trong m\u1ed9t chu k\u1ef3:<\/p>\n

C\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng c\u00f4ng su\u1ea5t theo \u0111\u1ecbnh ngh\u0129a: P = ((U_{0<\/sub>\u00a0.I0<\/sub>) \/ 2) .cos\u03c6 \u00a0\u00a0= U.I.cos\u03c6<\/p>\n}

2\/ C\u00e1c c\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t th\u01b0\u1eddng s\u1eed d\u1ee5ng:<\/h3>\n

P = U.I.cos\u1d69 \u00a0= I.Z.I.(R \/ Z) = R.I^{2<\/sup><\/p>\n}

\u21d2 P = R. I^{2<\/sup> = R.U2<\/sup> \/ (R2<\/sup> + (Z_{L<\/sub> – ZC<\/sub>)2<\/sup>)<\/p>\n}}

C\u00f4ng th\u1ee9c \u0111\u1eb7c bi\u1ec7t:<\/strong><\/p>\n

P = U.I.cos\u03c6 \u00a0 \u00a0= U (U\/Z)(R\/R). cos\u03c6 \u00a0= (U^{2<\/sup> \/ R).cos2<\/sup> \u03c6<\/p>\n}

\u21d2 P = (U^{2<\/sup> \/ R).cos2<\/sup> \u03c6<\/p>\n}

* H\u1ec7 s\u1ed1 c\u00f4ng su\u1ea5t:<\/p>\n

<\/p>\n

Ch\u00fa \u00fd: – \u03c0 \/ 2\u00a0\u2264\u00a0\u03c6 = \u03c6_{u<\/sub> – \u03c6i<\/sub>\u00a0\u2264\u00a0\u03c0 \/ 2\u00a0\u21d2 ta c\u00f3 0\u00a0\u2264\u00a0cos \u03c6\u00a0\u2264 1<\/p>\n}

II\/ V\u00ed d\u1ee5 \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t<\/h2>\n

V\u00ed d\u1ee5 1:\u00a0<\/strong>Cho \u0111i\u1ec7n \u00e1p\u00a0\u00a0(V)\u00a0v\u00e0o hai \u0111\u1ea7u c\u1ee7a \u0111o\u1ea1n m\u1ea1ch th\u00ec ta c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ea1ch l\u00e0\u00a0\u00a0(A). H\u00e3y t\u00ednh cos \u03c6 v\u00e0 P<\/p>\n

Gi\u1ea3i<\/strong><\/p>\n

Theo \u0111\u1ec1 b\u00e0i ta c\u00f3 :<\/p>\n

<\/p>\n

\u21d2\u00a0\u03c6 =\u00a0\u03c6_{u<\/sub> – \u03c6i<\/sub>\u00a0=\u00a0(- \u03c0 \/\u00a03) – 0 =\u00a0– \u03c0 \/ 3\u00a0\u21d2\u00a0cos \u03c6 =\u00a0\u00bd<\/p>\n}

\u21d2 C\u00f4ng su\u1ea5t c\u1ee7a d\u00f2ng \u0111i\u1ec7n l\u00e0\u00a0<\/p>\n

V\u00ed d\u1ee5 2:\u00a0<\/strong>Cho 1 \u0111i\u1ec7n \u00e1p\u00a0\u00a0(V) v\u00e0o hai \u0111\u1ea7u c\u1ee7a \u0111o\u1ea1n m\u1ea1ch RLC \u0111\u01b0\u1ee3c gh\u00e9p n\u1ed1i ti\u1ebfp nhau c\u00f3 R = 30(\u03a9), L = 3 \/ 5\u03c0 (H), C =\u00a010 ^{-3<\/sup> \/ 9\u03c0 (F). T\u00ednh P v\u00e0\u00a0 cos \u03c6<\/p>\n}

Gi\u1ea3i:\u00a0<\/strong><\/p>\n

Ta c\u00f3 Z_{L<\/sub> = Zw<\/sub> = (3 \/ 5\u03c0). 100\u03c0 = 60 (\u03a9)<\/p>\n}

Z_{C<\/sub> = 1 \/ Lw<\/sub> = 1\/ (10 ^{-3<\/sup> \/ 9\u03c0) . 100\u03c0 = 90\u00a0(\u03a9)<\/p>\n}}

\u21d2 P =\u00a0R.U^{2<\/sup> \/ (R2<\/sup> + (Z_{L<\/sub> – ZC<\/sub>)2<\/sup>) =\u00a030.1202<\/sup> \/ (302<\/sup> + (60\u00a0– 90)2<\/sup>) = 240 (W)<\/p>\n}}

\u21d2\u00a0<\/p>\n

V\u00ed d\u1ee5 3:\u00a0<\/strong>\u0110\u1eb7t m\u1ed9t \u0111i\u1ec7n \u00e1p xoay chi\u1ec1u v\u1edbi gi\u00e1 tr\u1ecb hi\u1ec7u d\u1ee5ng l\u00e0 100 V v\u00e0o hai \u0111\u1ea7u c\u1ee7a m\u1ea1ch RLC \u0111\u01b0\u1ee3c gh\u00e9p n\u1ed1i ti\u1ebfp c\u00f3 \u00a0R = 50(\u03a9)\u00a0th\u00ec \u0111i\u1ec7n \u00e1p 2 \u0111\u1ea7u c\u1ee7a m\u1ea1ch l\u1ec7ch pha l\u00e0 \u00a0\u00a0so v\u1edbi \u0111i\u1ec7n \u00e1p c\u1ee7a hai \u0111\u1ea7u t\u1ee5 C. T\u00ednh P, cos\u00a0\u03c6<\/p>\n

Gi\u1ea3i:<\/strong><\/p>\n

<\/p>\n

Ta c\u00f3 c\u00f4ng th\u1ee9c:<\/p>\n

P = (U^{2<\/sup> \/ R).cos2<\/sup> \u03c6<\/p>\n}

\u21d2\u00a0P = (100^{2<\/sup> \/ 50) . cos2<\/sup> (- \u03c0 \/ 6) = 150 (W)<\/p>\n}

\u21d2 ta c\u00f3\u00a0<\/p>\n

Hi v\u1ecdng v\u1edbi nh\u1eefng v\u00ed d\u1ee5 tr\u00ean c\u00e1c b\u1ea1n c\u00f3 th\u1ec3 gi\u1ea3i quy\u1ebft \u0111\u01b0\u1ee3c nh\u1eefng b\u00e0i to\u00e1n v\u1ec1 c\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t c\u1ee7a d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u<\/p>\n","protected":false},"excerpt":{"rendered":"

Trong ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc v\u1eadt l\u00fd l\u1edbp 12 th\u00ec chuy\u00ean \u0111\u1ec1 v\u1ec1 c\u00f4ng su\u1ea5t c\u1ee7a d\u00f2ng \u0111i\u1ec7n xoay chi\u1ec1u l\u00e0 kh\u00e1 quan tr\u1ecdng. Sau \u0111\u00e2y c\u00e1c b\u1ea1n h\u00e3y c\u00f9ng tamcaotrithuc.com… <\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[450,470],"tags":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/10451"}],"collection":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/comments?post=10451"}],"version-history":[{"count":0,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/10451\/revisions"}],"wp:attachment":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/media?parent=10451"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/categories?post=10451"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/tags?post=10451"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}