{"id":9154,"date":"2018-09-25T06:37:54","date_gmt":"2018-09-24T23:37:54","guid":{"rendered":"https:\/\/example.com\/?p=9154"},"modified":"2023-12-19T07:40:12","modified_gmt":"2023-12-19T00:40:12","slug":"latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12","status":"publish","type":"post","link":"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12","title":{"rendered":"Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12"},"content":{"rendered":"<p style=\"text-align: justify;\">Halo Squad,&nbsp;<span>yang berada di jenjang SMA dan mengambil jurusan IPA! Selamat menghadapi minggu UTS ya.&nbsp;Untuk membantu kamu mempersiapkan diri, kali ini Ruangguru sudah mempersiapkan latihan dan pembahasan soal UTS Fisika kelas 12.&nbsp;Yuk<span>, mari kita bahas.<\/span><\/span><\/p>\n<p><!--more--><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #cc0201;\"><strong>Topik: Listrik Arus Searah<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"font-weight: 400;\">1. Sebuah rangkaian listrik dengan sumber tegangan V memiliki kuat arus 6 A. Jika hambatan dibuat tetap, sedangkan sumber tegangan dinaikkan menjadi 2V, maka kuat arus akan menjadi\u2026<\/span><br \/><\/strong><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">12 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">3 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">1,5 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">24 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak berubah<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Berdasarkan hukum Ohm,<\/span><\/p>\n<p style=\"text-align: justify;\"><em><span style=\"font-weight: 400;\">V= I \u00d7 R&nbsp;<\/span><span style=\"font-weight: 400;\">.<\/span><\/em><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Dengan nilai <\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\"> yang tidak berubah maka,<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis1-10.png\" alt=\"fis1-10\" width=\"80\" style=\"width: 80px;\"><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/strong><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">2. Perhatikan pernyataan berikut!<\/span><\/strong><br \/><\/span><\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Menyala lebih terang<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Menyala lebih redup<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Jika salah satu lampu dicabut, lampu lain tetap menyala<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Jika salah satu lampu dicabut, lampu lainnya mati<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Terdapat dua buah rangkaian berbeda yang dihubungkan ke sebuah baterai dengan nilai tegangan yang sama. Pada rangkaian pertama, lampu A-B-C dipasang secara paralel sedangkan pada rangkaian kedua lampu D-E-F dipasang secara seri. Sifat di atas yang merupakan sifat rangkaian lampu A-B-C jika dibandingkan dengan lampu D-E-F adalah\u2026<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">1 dan 3<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2 dan 4<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">1 dan 4<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2 dan 3<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak ada yang benar<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Redup tidaknya lampu ditentukan oleh besarnya daya yang dikonsumsi oleh lampu. Jika keenam lampu dianggap memiliki nilai resistansi yang sama sebesar <\/span><span style=\"font-weight: 400;\">R<\/span><span style=\"font-weight: 400;\"> maka sejak kedua rangkaian dihubungkan dengan sumber tegangan yang sama <\/span><span style=\"font-weight: 400;\">V<\/span><span style=\"font-weight: 400;\">, maka<\/span><\/p>\n<p><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis24-1.png\" alt=\"fis24-1\" width=\"600\" style=\"width: 600px; display: block; margin: 0px auto;\"><br \/><span style=\"background-color: transparent;\"><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"background-color: transparent;\"><span style=\"color: #ff0201;\">Baca juga:<\/span>&nbsp;&nbsp;<\/span><span style=\"color: #3574e3;\"><a href=\"\/latihan-soal-uts-bahasa-inggris-kelas-12\" style=\"color: #3574e3;\"><span style=\"background-color: transparent;\">Latihan Soal UTS Bahasa Inggris Kelas 12<\/span><\/a><\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"background-color: transparent;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span style=\"background-color: transparent;\">3. Perhatikan gambar di bawah ini!<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis3-4.png\" alt=\"fis3-4\" width=\"200\" style=\"width: 200px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Jika saklar tetap berada dalam keadaan terbuka seperti pada gambar, lampu yang menyala adalah\u2026<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak ada<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">A dan B<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">C dan D<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">A, B, C, dan D<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak dapat ditentukan<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Lampu A dirangkai seri dengan lampu B, lampu C juga dirangkai seri dengan lampu D. Rangkaian lampu A-B dan lampu C-D dipasangkan paralel dengan sumber tegangan dan saklar. Maka, rangkaian lampu A-B dan lampu C-D mendapat sumber tegangan langsung dari sumber tegangan (karena dirangkaikan secara paralel) dan dialiri arus listrik karena rangkaiannya tidak terputus. Keberadaan saklar tidak berpengaruh pada rangkaian lampu manapun. Maka, lampu yang menyala adalah semua lampu (lampu A, B, C, dan D).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">4.&nbsp;Perhatikan gambar di bawah ini!<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis4-1.png\" alt=\"fis4-1\" width=\"320\" style=\"width: 320px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Jika diketahui <img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis5-1.png\" alt=\"fis5-1\" width=\"132\" style=\"width: 132px;\"><\/span><span style=\"font-weight: 400;\">&nbsp;secara berturut-turut adalah 6 \u03a9, 12 \u03a9, dan 6 \u03a9, serta sumber tegangannya adalah 12 V, besar kuat arus I<\/span><span style=\"font-weight: 400;\">2<\/span><span style=\"font-weight: 400;\"> adalah\u2026<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">1 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">4 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">3 A<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">0,5 A<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Berdasarkan Hukum II Kirchoff<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis6-1.png\" alt=\"fis6-1\" width=\"134\" style=\"width: 134px;\"><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Rangkaian di atas dapat dibagi menjadi 2 loop seperti di bawah ini<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis7-1.png\" alt=\"fis7-1\" width=\"319\" style=\"width: 319px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Berdasarkan Hukum I Kirchoff<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis8-1.png\" alt=\"fis8-1\" width=\"369\" style=\"width: 369px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Pada loop 1 didapat persamaan <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis9-1.png\" alt=\"fis9-1\" width=\"370\" style=\"width: 370px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Pada loop 2 didapat persamaan<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis10-1.png\" alt=\"fis10-1\" width=\"379\" style=\"width: 379px;\"><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Substitusikan persamaan (1) ke persamaan (2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis11-1.png\" alt=\"fis11-1\" width=\"163\" style=\"width: 163px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Substitusikan ke persamaan (3)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis12-1.png\" alt=\"fis12-1\" width=\"143\" style=\"width: 143px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">5. Sebuah lampu memiliki spesifikasi 20 W, 220 V. Jika lampu dipasang pada tegangan 110 V, maka energi listrik yang terpakai dalam 1 jam adalah\u2026<\/span><\/strong><br \/><\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">72 kJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">36 kJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">44 kJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">22 kJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">18 kJ<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : E<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Daya (P) berbanding lurus dengan tegangan (V) kuadrat. Maka<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis13-1.png\" alt=\"fis13-1\" width=\"266\" style=\"width: 266px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #cc0201;\"><strong>Topik: Listrik Statis<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">6. Terdapat dua buah muatan dengan muatan masing-masing +3 \u00b5C dan <\/span><span style=\"font-weight: 400;\">&#8211;<\/span><span style=\"font-weight: 400;\">3 \u00b5C. Kedua muatan terpisah sejauh 3 cm. Besar dan jenis gaya Coulomb antara kedua muatan tersebut adalah\u2026<\/span><\/strong><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">90 N dan tolak menolak<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">90 N dan tarik menarik<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2.700 N dan tolak menolak<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2.700 N dan tarik menarik<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak ada jawaban yang benar<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Besar gaya Coulomb dicari dengan<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis14-1.png\" alt=\"fis14-1\" width=\"204\" style=\"width: 204px;\"><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Karena muatan berbeda tanda, positif dan negaif, maka gaya yang bekerja tarik menarik.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">7.&nbsp;<\/span><strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">Dua buah muatan yang memiliki jarak R memiliki gaya Coulomb sebesar F. Jika jarak kedua muatan diubah menjadi 2R, gaya Coulomb antara kedua muatan tersebut menjadi\u2026<\/span><\/strong><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">2F<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">F\/2<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">4F<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">F\/4<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Tidak berubah<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Gaya Coulomb dirumuskan sebagai<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis15-1.png\" alt=\"fis15-1\" width=\"415\" style=\"width: 415px;\"><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/p>\n<p style=\"text-align: justify;\">8.&nbsp;<strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">Perhatikan gambar di bawah ini!<\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis16-1.png\" alt=\"fis16-1\" width=\"456\" style=\"width: 456px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Muatan A dan B terpisah sejauh 1,5 m seperti gambar di atas. Sebuah titik di antara kedua muatan tersebut memiliki medan listrik sama dengan nol. Jika muatan A bernilai +3 C dan muatan B bernilai +12 C, jarak titik tersebut dari muatan A adalah\u2026<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">0,3 m<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">1 m<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">0,5 m<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">0,6 m<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">0,9 m<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Medan listrik pada suatu titik dinyatakan<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis17-1.png\" alt=\"fis17-1\" width=\"63\" style=\"width: 63px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Jika di titik tersebut medan listriknya sama dengan nol, maka medan listrik A dan B di titik tersebut besarnya sama maka<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis18-1.png\" alt=\"fis18-1\" width=\"88\" style=\"width: 88px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Jarak titik tersebut ke muatan A ditambah jarak titik tersbut ke muatan B adalah 1,5 m. Maka<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis19-1.png\" alt=\"fis19-1\" width=\"160\" style=\"width: 160px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Maka, jarak titik tersebut ke muatan A adalah 0,5 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">9.&nbsp;Perhatikan gambar di bawah ini!<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis16-1.png\" alt=\"fis16-1\" width=\"456\" style=\"width: 456px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Sebuah titik terletak tepat di tengah muatan A dan B. Jika muatan A = + 0,2 nC dan muatan B = <\/span><span style=\"font-weight: 400;\">&#8211;<\/span><span style=\"font-weight: 400;\">0,3 nC, serta jarak A dan B adalah 6 cm, maka potensial listrik di titik tersebut adalah\u2026<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">180 V<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">150 V<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">-150 V<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">30 V<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">-30 V<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : E<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Titik tersebut mendapatkan pengaruh potensial listrik dari muatan A dan B. Titik berada tepat di tengah, sehingga jarak titik ke muatan A dan B sama, yaitu 6 cm dibagi 2, adalah 3 cm. Potensial listrik di titik tersebut adalah penjumlahan potensial listrik dari muatan A dan B. Potensial oleh sebuah muatan dinyatakan sebagai,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis20-1.png\" alt=\"fis20-1\" width=\"67\" style=\"width: 67px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Maka jika terdapat dua muatan A dan B,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis21-1.png\" alt=\"fis21-1\" width=\"324\" style=\"width: 324px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/strong><\/p>\n<p style=\"text-align: justify;\">10.&nbsp;<strong style=\"background-color: transparent;\"><span style=\"font-weight: 400;\">Terdapat rangkaian dengan dua buah kapasitor yang disusun seri. Jika kapasitansi kapasitor masing-masing adalah 6 \u03bcF dan 12 \u03bcF dan beda potensial rangkaian adalah 220 V, energi yang tersimpan adalah\u2026<\/span><\/strong><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">440,0 mJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">220,0 mJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">110,2 mJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">96.8 mJ<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">55,1 mJ<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban : D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan:<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Pada rangkaian seri, kapasitansi pengganti kapasitor adalah<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis22-1.png\" alt=\"fis22-1\" width=\"215\" style=\"width: 215px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\">Maka energi yang tersimpan pada rangkaian tersebut<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/fis23-1.png\" alt=\"fis23-1\" width=\"381\" style=\"width: 381px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design(4)-6.png\" alt=\"Untitled design(4)-6\" width=\"820\" style=\"width: 820px;\"><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span><em>Nah<\/em>, itulah 10 soal untuk latihan persiapan Ujian Tengah Semester Fisika kelas 12. Semoga membantu ya! Kalau 10 soal masih belum cukup, kamu bisa mencoba mengasah kemampuan dengan<\/span><span>&nbsp;<\/span><strong><a href=\"https:\/\/uji.ruangguru.com\/\" rel=\"noopener\" target=\"_blank\">ruanguji<\/a><\/strong><span>.<span>&nbsp;<\/span>Yuk, cobain sekarang!<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-weight: 400;\"><span><a href=\"https:\/\/www.ruangguru.com\/ruanguji\" target=\"_blank\" style=\"text-align: center;\" class=\"rg-cta\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/cta\/4d67385e-8dfc-4cd3-8931-2ec9d1ca1c6d.jpeg\" width=\"820\" height=\"200\" alt=\"IDN CTA Blog ruanguji Ruangguru\" \/><\/a><\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Halo Squad,&nbsp;yang berada di jenjang SMA dan mengambil jurusan IPA! Selamat menghadapi minggu UTS ya.&nbsp;Untuk membantu kamu mempersiapkan diri, kali ini Ruangguru sudah mempersiapkan latihan dan pembahasan soal UTS Fisika kelas 12.&nbsp;Yuk, mari kita bahas.<\/p>\n","protected":false},"author":22,"featured_media":9154,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_knawatfibu_url":["https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Latihan%20Soal%20UTS%20Fisika%20Kelas%2012.png"],"_edit_last":["1"],"_edit_lock":["1702946412:1"]},"categories":[552,561],"tags":[79,76,37],"class_list":["post-9154","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-latihan-soal","category-sma-kelas-12","tag-bahasa-inggris-xii","tag-kelas-12","tag-sma"],"aioseo_notices":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12 - Belajar Gratis di Rumah Kapan Pun! | Blog Ruangguru<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12 - Belajar Gratis di Rumah Kapan Pun! | Blog Ruangguru\" \/>\n<meta property=\"og:description\" content=\"Halo Squad,&nbsp;yang berada di jenjang SMA dan mengambil jurusan IPA! Selamat menghadapi minggu UTS ya.&nbsp;Untuk membantu kamu mempersiapkan diri, kali ini Ruangguru sudah mempersiapkan latihan dan pembahasan soal UTS Fisika kelas 12.&nbsp;Yuk, mari kita bahas.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\" \/>\n<meta property=\"og:site_name\" content=\"Belajar Gratis di Rumah Kapan Pun! | Blog Ruangguru\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/ruanggurucom\/\" \/>\n<meta property=\"article:published_time\" content=\"2018-09-24T23:37:54+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-12-19T00:40:12+00:00\" \/>\n<meta name=\"author\" content=\"Shabrina Alfari\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ruangguru\" \/>\n<meta name=\"twitter:site\" content=\"@ruangguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Shabrina Alfari\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"4 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12#article\",\"isPartOf\":{\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\"},\"author\":{\"name\":\"Shabrina Alfari\",\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/#\/schema\/person\/af6350034b171a1408a571ed11ae0248\"},\"headline\":\"Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12\",\"datePublished\":\"2018-09-24T23:37:54+00:00\",\"dateModified\":\"2023-12-19T00:40:12+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\"},\"wordCount\":850,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12#primaryimage\"},\"thumbnailUrl\":\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Latihan%20Soal%20UTS%20Fisika%20Kelas%2012.png\",\"keywords\":[\"Bahasa Inggris XII\",\"Kelas 12\",\"SMA\"],\"articleSection\":[\"Latihan Soal\",\"Latihan Soal SMA Kelas 12\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\",\"url\":\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\",\"name\":\"Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12 - Belajar Gratis di Rumah Kapan Pun! 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