{"id":134642,"date":"2017-07-05T16:41:21","date_gmt":"2017-07-05T09:41:21","guid":{"rendered":"https:\/\/video.quipper.com\/id\/blog\/?p=134642"},"modified":"2021-11-24T20:58:52","modified_gmt":"2021-11-24T13:58:52","slug":"anak-smp-yuk-buang-anggapan-fisika-itu-sulit-dengan-belajar-listrik-dinamis-kelas-9-lanjutan-dan-pembahasan-ini","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/fisika-un-sma\/anak-smp-yuk-buang-anggapan-fisika-itu-sulit-dengan-belajar-listrik-dinamis-kelas-9-lanjutan-dan-pembahasan-ini\/","title":{"rendered":"Anak SMP? Yuk Buang Anggapan Fisika Itu Sulit dengan Belajar Listrik Dinamis Kelas 9 Lanjutan dan Pembahasan Ini!"},"content":{"rendered":"

\"Anak<\/a><\/p>\n

Quipperian! Jika kita kemarin telah belajar mengenai Listrik Dinamis Kelas 9, maka kali ini kita akan belajar dari contoh soal Listrik Dinamis lanjutan lengkap dengan pembahasan! Yuk, kembali semangat dan hilangkan stigma Fisika itu sulit!<\/span><\/p>\n

Soal Listrik Dinamis Lanjutan 1<\/b><\/h2>\n

Budi membeli lampu bertuliskan 220 volt, 100 watt, perkirakan hambatan yang dimiliki lampu tersebut!<\/span><\/p>\n

Pembahasan<\/span><\/p>\n

Mencari hambatan dari daya lampu yang diketahui:<\/span><\/p>\n

P = V2\/ R<\/span><\/p>\n

R = V2\/ P<\/span><\/p>\n

R = 2202 \/ 100<\/span><\/p>\n

R = 484 O<\/span><\/p>\n

Soal Listrik Dinamis Lanjutan 2<\/b><\/h2>\n

Anto membeli lampu dengan spesifikasi 220 volt, 100 watt. Jika lampu dipasang Anto pada tegangan 110 volt, berapakah daya lampu saat menyala?<\/span><\/p>\n

Pembahasan<\/b><\/p>\n

Hambatan lampu saat dipasang pada tegangan 220 volt maupun 110 adalah tetap<\/span><\/p>\n

R2 = R1<\/span><\/p>\n

V22\/ P2 = V12\/ P1<\/span><\/p>\n

P2 = (V1\/V2)2 x P1<\/span><\/p>\n

P2 = (110\/220) 2 x 100 = 1\/4 x 100 watt = 25 watt<\/span><\/p>\n

Mau Lulus USBN? Ini Kisi-Kisi USBN Fisika 2017!<\/a><\/h3>\n

Soal Listrik Dinamis Lanjutan 3<\/b><\/h2>\n

 <\/p>\n

Pemanas air dengan spesifikasi 200 watt digunakan untuk memasak air bersuhu 30\u00b0C sebanyak 0,5 kg. Jika kalor jenis air adalah 4200 J\/kg\u00b0C tentukan berapa lama waktu yang diperlukan hingga air mencapai suhu 100\u00b0C !<\/span><\/p>\n

 <\/p>\n

Pembahasan<\/b><\/p>\n

Energi listrik diubah menjadi kalor untuk memanaskan air<\/span><\/p>\n

Data soal:<\/span><\/p>\n

P = 200 watt<\/span><\/p>\n

? T = 100 – 30 = 70 \u00b0C<\/span><\/p>\n

m = 0,5 kg<\/span><\/p>\n

c = 4200 J\/kg\u00b0C<\/span><\/p>\n

t = ……….<\/span><\/p>\n

 <\/p>\n

P x t = m x c x ?T<\/span><\/p>\n

200 x t = 0,5 x 4200 x 70<\/span><\/p>\n

t = 147000 \/ 200<\/span><\/p>\n

t = 735 detik<\/span><\/p>\n

t = 735 \/ 60 menit = 12,25 menit<\/span><\/p>\n

 <\/p>\n

Soal Listrik Dinamis Lanjutan 4<\/b><\/h2>\n

Sebuah rumah memasang 5 lampu 20 watt dan menyala 12 jam sehari, 2 lampu 60 watt menyala 5 jam sehari, sebuah kulkas 125 watt menyala 24 jam sehari, pesawat TV 200 watt menyala 6 jam sehari dan sebuah setrika listrik 250 watt yang dipakai 2 jam sehari. Jika tarif listrik adalah Rp 750 \/ kwh, perkirakan biaya listrik rumah tersebut dalam satu hari, satu minggu (7 hari) dan satu bulan (30 hari)!<\/span><\/p>\n

Pembahasan<\/span><\/p>\n

Cara menghitung tarif listrik :<\/span><\/p>\n

Pola:<\/span><\/p>\n

(watt alat)x(jumlah alat)x(lama pemakaian)<\/span><\/p>\n

Lampu ? 20 watt x 5 x 12 jam = 1200 wh<\/span><\/p>\n

Lampu ? 60 watt x 2 x 5 jam = 600 wh<\/span><\/p>\n

Kulkas ? 125 watt x 1 x 24 jam = 3000 wh<\/span><\/p>\n

TV 200 ? watt x 1 x 6 jam = 1200 wh<\/span><\/p>\n

Setrika ? 250 watt x `1 x 2 jam = 500 wh<\/span><\/p>\n

——————————————————- +<\/span><\/p>\n

Energi dipakai dalam sehari = 6500 wh = 6,5 kwh<\/span><\/p>\n

Tarif listrik dalam satu hari = Rp 750 x 6,5 = Rp 4875<\/span><\/p>\n

Tarif listrik dalam satu minggu = Rp 4875 x 7 = Rp 34125<\/span><\/p>\n

Tarif listrik dalam satu bulan = Rp 4875 x 30 = 146250<\/span><\/p>\n

Mau Lulus USBN? Ini Kisi-Kisi USBN Fisika 2017!<\/a><\/h3>\n

Soal Listrik Dinamis Lanjutan 5<\/b><\/h2>\n

Jika: <\/span><\/p>\n

R1 = R2 = 10 O<\/span><\/p>\n

R3 = R4 = 8 O<\/span><\/p>\n

Berapakah besarnya kuat arus (I) yang mengalir?<\/span><\/p>\n

    \n
  1. 0,5 A<\/span><\/li>\n
  2. 2 A<\/span><\/li>\n
  3. 36 A<\/span><\/li>\n
  4. 288 A<\/span><\/li>\n<\/ol>\n

    (Dari Soal UN IPA 2008\/2009)<\/span><\/p>\n

    Pembahasan<\/b><\/p>\n

    Tentukan hambatan pengganti atau hambatan totalnya terlebih dulu:<\/span><\/p>\n

    R1 dan R2 susunan seri, tinggal dijumlahkan saja<\/span><\/p>\n

    R12 = 10 + 10 = 20 O<\/span><\/p>\n

    R3 dan R4 susunan paralel, pakai seper-seper terus dibalik:<\/span><\/p>\n

    1\/R34 = 1\/8 + 1\/8 = 2\/8<\/span><\/p>\n

    R34 = 8\/2 = 4 O<\/span><\/p>\n

    R12 dan R34 lantas diseri lagi biar jadi Rtotal<\/span><\/p>\n

    Rtotal = 20 + 4 = 24 O<\/span><\/p>\n

    Terakhir menghitung arus rangkaian:<\/span><\/p>\n

    I = V\/R = 12\/24 = 0,5 A<\/span><\/p>\n

    Penulis: Sritopia<\/b><\/p>\n","protected":false},"excerpt":{"rendered":"

    Quipperian! Jika kita kemarin telah belajar mengenai Listrik Dinamis Kelas 9, maka kali ini kita akan belajar dari contoh soal Listrik Dinamis lanjutan lengkap…<\/p>\n","protected":false},"author":156446345,"featured_media":134652,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_wpcom_ai_launchpad_first_post":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[677999187],"tags":[],"ppma_author":[679386822],"class_list":["post-134642","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-fisika-un-sma"],"acf":[],"yoast_head":"\nAnak SMP? 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