{"id":11479,"date":"2024-11-12T15:43:42","date_gmt":"2024-11-12T15:43:42","guid":{"rendered":"https:\/\/fisicaevestibular.com.br\/novo\/?page_id=11479"},"modified":"2024-12-03T14:11:13","modified_gmt":"2024-12-03T14:11:13","slug":"caracteristicas-das-associacoes-serie-paralelo-resolucao-en","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/eletrodinamica-caracteristicas-das-associacoes-serie-paralelo\/caracteristicas-das-associacoes-serie-paralelo-resolucao-en\/","title":{"rendered":"Characteristics of Series-Parallel Associations – Resolution – EN"},"content":{"rendered":"

CHARACTERISTICS OF SERIES-PARALLEL ASSOCIATIONS \u2013 RESOLUTION<\/span><\/h1>\n
\n

Commented Resolution<\/span><\/b><\/b><\/p>\n

Characteristics of Series-Parallel Associations<\/span><\/b><\/p>\n

 <\/p>\n

 <\/p>\n

01-(ENEM-MEC)<\/span><\/b><\/p>\n

Calculation of the electric current i of each fuse using the\u00a0<\/span><\/b>nominal values \u200b\u200bP=iU \u2014 i=55\/36 A \u2014 when connected to the same fuse, the new current i’ passes through the two lamps, each of<\/span><\/b><\/p>\n

\"\"<\/p>\n

resistance R, are associated in parallel and, in this case, the equivalent resistor is worth Req=R\/2 \u2014 if the equivalent resistance drops by half, the new current i’ is equal to twice the previous one, since, R=U\/i with constant U \u2014 or, both lamps are working within specifications and therefore run by current i=55\/36 A \u2014 i’=2.(55\/36) \u2014 i’=3.05 A \u2014 to support this electric current, the lowest value of the fuse must be 5 A, that is, orange \u2014\u00a0\u00a0\u00a0<\/span><\/b>R-C<\/span><\/b><\/p>\n

 <\/p>\n

02-(FATEC-SP)<\/span><\/b><\/p>\n

Equivalent resistance of m resistors of electrical resistance R1 \u2014 Req1=mR1 \u2014 idem \u2014 Req2=nR2 \u2014 Reqtotal=mR1 + nR2 \u2014 the current is the same in all resistors (series association) and is worth \u2014 Reqtotal=U\/i \u2014 (mR1 + nR2)=U\/i \u2014 i=U\/(mR1 +nR2) \u2014\u00a0\u00a0<\/span><\/b>R- A<\/span><\/b><\/p>\n

03-(CEFET-RJ)<\/span><\/b><\/p>\n

The total voltage is the sum of the voltages of each lamp \u2014 U\u00a0<\/span><\/b>t<\/span><\/b><\/sub>\u00a0= nU\u00a0<\/span><\/b>l<\/span><\/b><\/sub>\u00a0\u00a0 \u2014 117 = n.15 \u2014 n = 117\/15 \u2014 n = 7.8 \u2014 as the voltage at the terminals of each lamp cannot exceed 15V, 8 lamps must be selected \u2014\u00a0\u00a0<\/span><\/b>R- D<\/b><\/span>\u00a0<\/b><\/b><\/p>\n

04-(FUVEST-SP)<\/span><\/b><\/p>\n

Closing switch C causes a short circuit in the terminals of lamp A, the current diverts and it goes out. Thus, as the total resistance decreases, the current increases in lamp B, increasing its brightness.\u00a0\u00a0<\/span><\/b><\/p>\n

A- A<\/span><\/b><\/p>\n

05-(UNIFESP-SP)<\/span><\/b><\/p>\n

Power supplied by the battery \u2014 Pb=iU=330.10-3.6 \u2014 Pb=1.98W \u2014 as the LED consumes a power of 1.0W, there is a power left for the resistor R of 1.98 \u2013 1.0 \u2014 PR=0.98W \u2014 PR=Ri\u00a0<\/span><\/b>2<\/span><\/b><\/sup>\u00a0\u00a0 \u20140.98=R.(330.10\u00a0<\/span><\/b>-3<\/span><\/b><\/sup>\u00a0)\u00a0<\/span><\/b>2<\/span><\/b><\/sup>\u00a0\u00a0 \u2014 R=98.10\u00a0<\/span><\/b>-2<\/span><\/b><\/sup>\u00a0\/(33.10\u00a0<\/span><\/b>-2<\/span><\/b><\/sup>\u00a0)\u00a0<\/span><\/b>2<\/span><\/b><\/sup>\u00a0\u00a0 \u2014 R=9.0\u03a9 \u2014\u00a0\u00a0\u00a0<\/span><\/b>R- C<\/span><\/b><\/p>\n

06-(PUC-SP)<\/span><\/b><\/p>\n

Correct answer \u2014 circuit in figure III where you can turn on the lamp, independently, with any of the switches \u2014 observe the sequence below:<\/span><\/b><\/p>\n

\"\"<\/p>\n

R-C.<\/span><\/b><\/p>\n

07-(FGV-SP)<\/span><\/b><\/p>\n

I- False \u2014 when one of the bulbs burns out, in the series circuit, the electric current is interrupted and the other bulb goes out.<\/span><\/b><\/p>\n

II- False \u2014 as the lamps are identical, each one will be subject to a voltage of 110V and the power in each one will be 4 times lower, that is, 25W.<\/span><\/b><\/p>\n

III- False \u2014 as the lamps are identical and each of them will be subject to a voltage of 110V, they will be within specifications and functioning normally.<\/span><\/b><\/p>\n

IV- Correct \u2014 as they are identical, each of them will have half of the total voltage, which is 220V, that is, each one will be subject to a voltage of 110V.<\/span><\/b><\/p>\n

R-B<\/span><\/b><\/p>\n

08-(PUC-SP)<\/span><\/b><\/p>\n

Series \u2014 R1 + R2=6 \u2014 R1=6 \u2013 R2 \u2014 parallel \u2014 R1.R2\/(R1 + R2)=4\/3 \u2014 (6 \u2013 R2).R2=(6 \u2013 R2 + R2) \u2014 R2\u00a0<\/span><\/b>2<\/span><\/b><\/sup>\u00a0\u00a0\u2013 6R2 + 8=0 \u2014 \u221a\u0394=2 \u2014 R2=(6 \u00b1 2)\/2 \u2014 R’2=4\u03a9 \u2014 R”=2\u03a9 \u2014 when one is 2\u03a9, the other is 4\u03a9 and vice versa \u2014\u00a0\u00a0\u00a0<\/span><\/b>R- C<\/span><\/b><\/p>\n

09-(UFPE-PE)<\/span><\/b><\/p>\n

R and the lamp are in parallel and both under ddp of 12V \u2014 lamp \u2014 P=iU \u2014 6=il.12 \u2014 il=0.5A \u2014 Rl=U\/i=12\/0.5 \u2014 Rl=24\u03a9 \u2014 Req= 24R\/(24 + R) \u2014 Req=U\/i \u2014 24R\/(24 + R)=12\/3 \u2014 20R=96 \u2014 R=4.8\u03a9 \u2014\u00a0\u00a0\u00a0<\/span><\/b>R- E<\/span><\/b><\/p>\n

10-(UCS-RS)<\/span><\/b><\/p>\n

The upper wire corresponds to one of the poles (phases) of the source and the lower one to the other \u2014 observe the figures carefully \u2014\u00a0\u00a0\u00a0<\/span><\/b>R- E<\/span><\/b><\/p>\n

11-(FUVEST-SP)<\/span><\/b><\/p>\n

Current in the lamp \u2014 Pl=il.U \u2014 150=il.110 \u2014 il=214\/15A \u2014 current in the iron \u2014 as the lamp and iron are associated in parallel \u2014 idisbreaker=il + if \u2014 if=(15 \u2013 214\/15) \u2014 if=214\/15A \u2014 Pf=if.U \u2014 Pf=214\/15.110=1,570W \u2014\u00a0\u00a0\u00a0<\/span><\/b>R- B<\/span><\/b><\/p>\n

 <\/p>\n

Back to exercises<\/span><\/a><\/h3>\n<\/div>\n

\n","protected":false},"excerpt":{"rendered":"

CHARACTERISTICS OF SERIES-PARALLEL ASSOCIATIONS \u2013 RESOLUTION Commented Resolution Characteristics of Series-Parallel Associations     01-(ENEM-MEC) Calculation of the electric current i of each fuse using the\u00a0nominal values \u200b\u200bP=iU \u2014 i=55\/36 A \u2014 when connected to the same fuse, the new current i’ passes through the two lamps, each of resistance R, are associated in parallel and, in this case, the<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3798,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-11479","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11479","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=11479"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11479\/revisions"}],"predecessor-version":[{"id":12057,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/11479\/revisions\/12057"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3798"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=11479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}