{"id":1725,"date":"2015-12-31T03:07:45","date_gmt":"2015-12-31T03:07:45","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1725"},"modified":"2024-11-08T17:00:56","modified_gmt":"2024-11-08T17:00:56","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-resistores-primeira-lei-de-ohm","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/eletricidade\/eletrodinamica\/resistores-primeira-lei-de-ohm\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-resistores-primeira-lei-de-ohm\/","title":{"rendered":"Primeira lei de Ohm \u2013 Resolu\u00e7\u00e3o EN"},"content":{"rendered":"

OHM’S FIRST LAW \u2013 RESOLUTION<\/h1>\n
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Commented resolution of entrance exam exercises on<\/b><\/b><\/p>\n

\u00a0Resistors<\/span><\/a>\u00a0\u2013 Ohm\u2019s First Law \u2013 Electrical Power of a\u00a0\u00a0Resistor<\/span><\/a><\/b><\/p>\n

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01\u00a0<\/b>\u2013<\/b>\u00a0RA \u2013 see theory<\/b><\/p>\n

02-<\/b>\u00a0\u00a0For i=2 A \u2014 U=5.i\u00a0<\/b>2<\/b><\/sup>\u00a0=5.2\u00a0<\/b>2<\/b><\/sup>\u00a0=5.4 \u2014 U=20V \u2014 R=U\/i=20\/2 \u200b\u200b\u2014 R=5 \u03a9 \u2014\u00a0\u00a0<\/b>\u00a0R- A<\/b><\/p>\n

03<\/b>\u00a0\u2013\u00a0<\/b>\u00a0In every electrical circuit there must be a generator (battery), electrical wires conducting the electrical current and a conductor (lamp) \u2014\u00a0<\/b>\u00a0 R- A\u00a0<\/b>.<\/b><\/p>\n

04-<\/b>\u00a0\u00a0R=U\/i \u2014 1,500=220\/i \u2014 i=220\/1,500 \u2014 i=0.1466 A \u2014 i=146.6mA \u2014\u00a0\u00a0<\/b>\u00a0R- D<\/b><\/p>\n

05-<\/b>\u00a0\u00a0a) Since the relationship between voltage (U) and current (i) is constant, the resistor is ohmic. Applying Ohm’s law \u2014 when t=10s \u2014 U=0.5V and i=1.0A \u2014 R=U\/i=0.5\/1.0 \u2014 R=0.5\u03a9<\/b><\/p>\n

b) The electrical power in the resistor is defined by the product Ui The area represents the dissipated energy.<\/b><\/p>\n

06-<\/b>\u00a0\u00a0R=U\/i \u2014 10\u00a0<\/b>3<\/b><\/sup>\u00a0=220\/i \u2014 i=220\/10\u00a0<\/b>3<\/b><\/sup>\u00a0\u00a0 \u2014 i=0.220 \u2014\u00a0\u00a0\u00a0<\/b>R- D<\/b><\/p>\n

07-<\/b>\u00a0\u00a0U=10,000V, i=300mA=0.3a \u2014 calculation of the total resistance so that the person does not die \u2014 R=U\/i=10,000\/0.3 \u2014 R=3.33.10\u00a0<\/b>5<\/b><\/sup>\u00a0\u03a9 \u2014 being the resistance of the person’s body R\u00a0<\/b>p<\/b><\/sub>\u00a0=2,000\u03a9, the electrical resistance of the air must complete this value \u2014 R=R\u00a0<\/b>p<\/b><\/sub>\u00a0\u00a0+ R\u00a0<\/b>air<\/b><\/sub>\u00a0\u00a0 \u2014\u00a0<\/b><\/p>\n

3.33.10\u00a0<\/b>5<\/b><\/sup>\u00a0=2,000 + R\u00a0<\/b>air<\/b><\/sub>\u00a0\u00a0 \u2014\u00a0\u00a0\u00a0<\/b>R\u00a0<\/b>air<\/b><\/sub>\u00a0=3.31.10\u00a0<\/b>5<\/b><\/sup>\u00a0\u03a9<\/b><\/p>\n

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