{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=4\u2126 v\u00e0 $R_{2}^{{}}$=12\u2126 m\u1eafc song song c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?","select":["A. 16\u2126 ","B. 48\u2126","C. 0,33\u2126","D. 3\u2126"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: <br\/>$R_{12}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{4\\times 12}{4+12}=3\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2771},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/h2.jpg'\/><\/center>Trong m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, hi\u1ec7u \u0111i\u1ec7n th\u1ebf U v\u00e0 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ \u0111\u01b0\u1ee3c gi\u1eef kh\u00f4ng \u0111\u1ed5i. H\u1ecfi khi gi\u1ea3m d\u1ea7n \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$ th\u00ec c\u01b0\u1eddng \u0111\u1ed9 I c\u1ee7a m\u1ea1ch \u0111i\u1ec7n ch\u00ednh s\u1ebd thay \u0111\u1ed5i nh\u01b0 th\u1ebf n\u00e0o? ","select":["A. T\u0103ng","B. Kh\u00f4ng thay \u0111\u1ed5i","C. Gi\u1ea3m","D.L\u00fac \u0111\u1ea7u t\u0103ng, sau \u0111\u00f3 gi\u1ea3m"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a m\u1ea1ch:<br\/>$R_{12}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}$<br\/>Khi gi\u1ea3m d\u1ea7n \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$ th\u00ec \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng gi\u1ea3m m\u00e0 $I=\\dfrac{U}{R}$ n\u00ean c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng di\u1ec7n t\u0103ng<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2772},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/h3.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 h\u00ecnh v\u1ebd, trong \u0111\u00f3 $R_{1}^{{}}$=5\u03a9, $R_{2}^{{}}$=10\u03a9, ampe k\u1ebf A1 ch\u1ec9 0,6A. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u AB c\u1ee7a \u0111o\u1ea1n m\u1ea1ch. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n \u1edf m\u1ea1ch ch\u00ednh.","select":["A. 0,3V; 0,9A","B. 0,9V;0,3A","C. 3V;9A","D. 3V;0,9A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch:<br\/>$R_{12}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{5\\times 10}{5+10}=\\dfrac{10}{3}\\Omega $<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: <br\/>$U=U_{1}^{{}}=I_{1}^{{}}\\times R=0,6\\times 5=3V$<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n \u1edf m\u1ea1ch ch\u00ednh:<br\/>$I=\\dfrac{U}{R_{12}^{{}}}=\\dfrac{3}{\\dfrac{10}{3}}=0,9A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2773},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho hai \u0111i\u1ec7n tr\u1edf, $R_{1}^{{}}$=15\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 t\u1ed1i \u0111a 2A v\u00e0 $R_{2}^{{}}$=10\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 t\u1ed1i \u0111a 1A. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a c\u00f3 th\u1ec3 \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ m\u1eafc song song l\u00e0:","select":["A. 40V","B. 10V","C. 30V","D. 25V"],"hint":"","explain":"<span class='basic_left'>Do $R_{2}^{{}}$=10\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 t\u1ed1i \u0111a 1A n\u00ean $I_{2}^{{}}$=1A<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a c\u00f3 th\u1ec3 \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m $R_{1}^{{}}$v\u00e0 $R_{2}^{{}}$ m\u1eafc song song l\u00e0:<br\/> $U=U_{2}^{{}}=I_{2}^{{}}\\times R_{2}^{{}}=1\\times 10=10V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2774},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=4\u2126 v\u00e0 $R_{2}^{{}}$=12\u2126 m\u1eafc song song c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?","select":["A. 16\u2126","B. 48\u2126","C. 0,33\u2126","D. 3\u2126"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=4\u2126 v\u00e0 $R_{2}^{{}}$=12\u2126 m\u1eafc song song c\u00f3 gi\u00e1 tr\u1ecb:<br\/>$R_{12}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{4\\times 12}{4+12}=3\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2775},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/h6.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 h\u00ecnh v\u1ebd, trong \u0111\u00f3 $R_{1}^{{}}$=20\u03a9, $R_{2}^{{}}$=30\u03a9, ampe k\u1ebf ch\u1ec9 1,2A. T\u00ednh s\u1ed1 ch\u1ec9 c\u1ee7a c\u00e1c ampe k\u1ebf $A_1$ v\u00e0 $A_2$. ","select":["A. 0,72A v\u00e0 0,48A","B. 0,56A v\u00e0 0,24A","C. 0,64A v\u00e0 0,44A","D. 0,72A v\u00e0 0,36A"],"hint":"","explain":"<span class='basic_left'>$R_{12}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{20\\times 30}{20+30}=12\\Omega $<br\/>V\u1eady $U_{AB}^{{}}=I_{{}}^{{}}\\times R_{AB}^{{}}=1,2\\times 12=14,4V$<br\/>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf 1: $I_{1}^{{}}=\\dfrac{U_{AB}^{{}}}{R_{1}^{{}}}=\\dfrac{4,4}{20}=0,72A$<br\/>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf 2: $I_{2}^{{}}=\\dfrac{U_{AB}^{{}}}{R_{2}^{{}}}=\\dfrac{4,4}{30}=0,48A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2776},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/h7.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 $R_{1}^{{}}$=15\u03a9, $R_{2}^{{}}$=10\u03a9, v\u00f4n k\u1ebf ch\u1ec9 12V. T\u00ednh s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf \u1edf m\u1ea1ch ch\u00ednh. ","select":["A. 1A","B. 2A","C. 3A","D. 4A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch:<br\/>$R_{AB}^{{}}=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{15\\times 10}{15+10}=6\\Omega $<br\/>$\\Rightarrow I_{AB}^{{}}=\\dfrac{U}{R_{AB}^{{}}}=\\dfrac{12}{6}=2A$<br\/>V\u1eady ampe k\u1ebf \u1edf m\u1ea1ch ch\u00ednh ch\u1ec9 2A<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2777},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/5.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, v\u00f4n k\u1ebf ch\u1ec9 36V, ampe k\u1ebf A ch\u1ec9 3A, $R_{1}^{{}}$=30\u03a9. T\u00ednh s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf A1.","select":["A. 2,4A","B. 1,5A","C. 1,2A","D. 1 A"],"hint":"","explain":"<span class='basic_left'>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf 1 l\u00e0:<br\/>$I_{1}^{{}}=\\dfrac{U}{R_{1}^{{}}}=\\dfrac{36}{30}=1,2A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2778},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=10\u03a9, $R_{2}^{{}}$=$R_{3}^{{}}$=20\u03a9 \u0111\u01b0\u1ee3c m\u1eafc song song v\u1edbi nhau v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 12V. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch.","select":["A. 12\u03a9","B. 8\u03a9","C. 10\u03a9","D. 5\u03a9"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch:<br\/>$R_{23}^{{}}=\\dfrac{R_{2}^{{}}R_{3}^{{}}}{R_{2}^{{}}+R_{3}^{{}}}=\\dfrac{20\\times 20}{20+20}=10\\Omega $<br\/>$\\Rightarrow R_{td}^{{}}=\\dfrac{R_{1}^{{}}R_{23}^{{}}}{R_{1}^{{}}+R_{23}^{{}}}=\\dfrac{10\\times 10}{10+10}=5\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2779},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ \u0111\u01b0\u1ee3c m\u1eafc song song v\u1edbi nhau, trong \u0111\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=6\u2126 ;d\u00f2ng \u0111i\u1ec7n m\u1ea1ch ch\u00ednh c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 I=1,2A v\u00e0 d\u00f2ng \u0111i\u1ec7n \u0111i qua \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$ c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 $I_{2}^{{}}$=0,4A. T\u00ednh $R_{2}^{{}}$.","select":["A. 10\u03a9","B. 12\u03a9","C. 15\u03a9","D. 13\u03a9"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $I_{1}^{{}}=I-I_{2}^{{}}=1,2-0,4=0,8A$<br\/>$U_{1}^{{}}=I_{1}^{{}}\\times R_{1}^{{}}=0,8\\times 6=4,8V$ <br\/>$\\Rightarrow U=U_{1}^{{}}=U_{2}^{{}}=4,8V$$\\Rightarrow R_{2}^{{}}=\\dfrac{U_{2}^{{}}}{I_{2}^{{}}}=12\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2780},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ = 4$R_{1}^{{}}$\u0111\u01b0\u1ee3c m\u1eafc song song v\u1edbi nhau. Khi t\u00ednh theo $R_{1}^{{}}$ th\u00ec \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch n\u00e0y c\u00f3 k\u1ebft qu\u1ea3 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?","select":["A. 5$R_{1}^{{}}$","B. 4$R_{1}^{{}}$","C. 0,8$R_{1}^{{}}$","D. 1,25$R_{1}^{{}}$"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng:<br\/> $R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{R_{1}^{{}}\\times 4R_{1}^{{}}}{R_{1}^{{}}+4R_{1}^{{}}}=0,8R_{1}^{{}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2781},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Ba \u0111i\u1ec3m tr\u1edf $R_{1}^{{}}$ = 5\u03a9, $R_{2}^{{}}$ = 10\u03a9 v\u00e0 $R_{3}^{{}}$ = 30\u03a9 \u0111\u01b0\u1ee3c m\u1eafc song song v\u1edbi nhau. \u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch song song n\u00e0y l\u00e0 bao nhi\u00eau?","select":["A. 0,33\u03a9","B. 3\u03a9","C. 33,3\u03a9","D. 45\u03a9"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $R_{12}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{5\\times 10}{5+10}=\\dfrac{10}{3}\\Omega $<br\/>$\\Rightarrow R_{123}^{{}}=\\dfrac{R_{12}^{{}}\\times R_{3}^{{}}}{R_{12}^{{}}+R_{3}^{{}}}=\\dfrac{\\dfrac{10}{3}\\times 30}{\\dfrac{10}{3}+30}=3\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2782},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 1,8V v\u00e0 hai \u0111i\u1ec7n tr\u1edf $R_{1}$ v\u00e0 $R_{2}$. N\u1ebfu m\u1eafc n\u1ed1i ti\u1ebfp hai \u0111i\u1ec7n tr\u1edf n\u00e0y v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U th\u00ec d\u00f2ng \u0111i\u1ec7n \u0111i qua ch\u00fang c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 $I_{1}^{{}}$ = 0,2A ; n\u1ebfu m\u1eafc song song hai \u0111i\u1ec7n tr\u1edf n\u00e0y v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U th\u00ec d\u00f2ng \u0111i\u1ec7n m\u1ea1ch ch\u00ednh c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 $I_{2}^{{}}$ = 0,9A. T\u00ednh $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$","select":["A. $R_{1}^{{}}$=1\u03a9 ; $R_{2}^{{}}$=2\u03a9","B. $R_{1}^{{}}$=2\u03a9 ; $R_{2}^{{}}$=4\u03a9","C. $R_{1}^{{}}$=3\u03a9 ; $R_{2}^{{}}$=6\u03a9","D. $R_{1}^{{}}$=4\u03a9 ; $R_{2}^{{}}$=8\u03a9"],"hint":"","explain":"<span class='basic_left'>Khi $R_{1}^{{}}$ n\u1ed1i ti\u1ebfp $R_{2}^{{}}$:<br\/>$R=R_{1}^{{}}+R_{2}^{{}}=\\dfrac{U}{I_{1}^{{}}}=\\dfrac{1,8}{0,2}=9\\Omega $ (1)<br\/>Khi $R_{1}^{{}}$ song song $R_{2}^{{}}$: $R=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{U}{I_{2}^{{}}}=\\dfrac{1,8}{0,9}=2\\Omega $ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta \u0111\u01b0\u1ee3c: $R_{1}^{{}}\\times R_{2}^{{}}=18\\Rightarrow R_{1}^{{}}=\\dfrac{18}{R_{2}^{{}}}$ <br\/>$\\Rightarrow R{{_{2}^{{}}}^{2}}-9R_{2}^{{}}+18=0$<br\/>Suy ra$R_{1}^{{}}$=3\u03a9 ; $R_{2}^{{}}$=6\u03a9 Ho\u1eb7c $R_{1}^{{}}$=6\u03a9 ; $R_{2}^{{}}$=3\u03a9 <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2783},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai5/lv2/img\/h14.jpg'\/><\/center>M\u1ed9t \u0111o\u1ea1n m\u1ea1ch g\u1ed3m ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 9\u03a9, $R_{2}^{{}}$ = 18\u03a9 v\u00e0 $R_{3}^{{}}$= 24\u03a9 \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 3,6V nh\u01b0 s\u01a1 \u0111\u1ed3 h\u00ecnh v\u1ebd. T\u00ednh s\u1ed1 ch\u1ec9 I c\u1ee7a ampe k\u1ebf A v\u00e0 s\u1ed1 ch\u1ec9 $I_{12}^{{}}$c\u1ee7a ampe k\u1ebf $A_1$.","select":["A. $I_{1}^{{}}$= 0,25A ; $I_{12}^{{}}$= 0,2A","B. $I_{1}^{{}}$= 0,5A ; $I_{12}^{{}}$= 0,4A","C. $I_{1}^{{}}$= 0,75A ; $I_{12}^{{}}$= 0,6A","D. $I_{1}^{{}}$= 1A ; $I_{12}^{{}}$= 0,8A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: <br\/>$\\dfrac{1}{R_{td}^{{}}}=\\dfrac{1}{R_{1}^{{}}}+\\dfrac{1}{R_{2}^{{}}}+\\dfrac{1}{R_{3}^{{}}}=\\dfrac{1}{9}+\\dfrac{1}{18}+\\dfrac{1}{24}$ $\\Rightarrow R_{td}^{{}}=4,8\\Omega $<br\/>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf: <br\/>$I=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{3,6}{4,8}=0,75A$<br\/>$\\dfrac{1}{R_{12}^{{}}}=\\dfrac{1}{R_{1}^{{}}}+\\dfrac{1}{R_{2}^{{}}}=\\dfrac{1}{9}+\\dfrac{1}{18}$<br\/>$\\Rightarrow R_{12}^{{}}=6\\Omega \\Rightarrow I_{12}^{{}}=\\dfrac{U}{R_{12}^{{}}}=\\dfrac{3,6}{6}=0,6A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2784},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"M\u1ed9t \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 6\u03a9, $R_{2}^{{}}$ = 3\u03a9 m\u1eafc song song v\u1edbi nhau v\u00e0o hai \u0111i\u1ec3m c\u00f3 hi\u1ec7u \u0111i\u1ec7n th\u1ebf 6V. \u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u00e0 c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ea1ch ch\u00ednh l\u00e0:","select":["A. R = 9\u03a9 v\u00e0 I = 0,6A","B. R = 9\u03a9 v\u00e0 I = 1A","C. R = 2\u03a9 v\u00e0 I = 1A","D. R = 2\u03a9 v\u00e0 I = 3A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng: $R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{6\\times 3}{6+3}=2\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch ch\u00ednh: $I=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{6}{2}=3A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2785},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 30\u03a9; $R_{2}^{{}}$ = 60\u03a9. M\u1eafc $R_{1}^{{}}$ song song $R_{2}^{{}}$ v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 12V. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch ch\u00ednh l\u00e0:","select":["A. 1A","B. 0,6A","C. 2A","D. 0,5A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng: $R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{30\\times 60}{30+60}=20\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch ch\u00ednh: $I=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{12}{20}=0,6A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2786},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 2\u03a9; $R_{2}^{{}}$ = 3\u03a9; $R_{3}^{{}}$ = 6\u03a9 \u0111\u01b0\u1ee3c m\u1eafc song song gi\u1eefa hai \u0111i\u1ec3m A v\u00e0 B c\u00f3 hi\u1ec7u \u0111i\u1ec7n th\u1ebf $U_{AB}^{{}}$. Khi \u0111\u00f3 c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua $R_{1}^{{}}$ l\u00e0 2A. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua c\u00e1c \u0111i\u1ec7n tr\u1edf c\u00f2n l\u1ea1i v\u00e0 hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111i\u1ec3m AB.","select":["A. $U_{AB}^{{}}$= 1V ; $I_{2}^{{}}$=0,3A;$I_{3}^{{}}$=0,9A","B. $U_{AB}^{{}}$= 2V ; $I_{2}^{{}}$=0,2A;$I_{3}^{{}}$=0,66A ","C. $U_{AB}^{{}}$= 3V ; $I_{2}^{{}}$=0,1A;$I_{3}^{{}}$=0,33A","D. $U_{AB}^{{}}$= 4V ; $I_{2}^{{}}$$=\\dfrac{4}{3}A$;$I_{3}^{{}}$$=\\dfrac{2}{3}A$"],"hint":"","explain":"<span class='basic_left'>Do 3 \u0111i\u1ec7n tr\u1edf m\u1eafc song song n\u00ean:<br\/>$U_{AB}^{{}}=U_{1}^{{}}=U_{2}^{{}}=U_{3}^{{}}=I_{1}^{{}}\\times R_{1}^{{}}=2\\times 2=4V$<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua c\u00e1c \u0111i\u1ec7n tr\u1edf:<br\/>$I_{2}^{{}}=\\dfrac{U_{AB}^{{}}}{R_{2}^{{}}}=\\dfrac{4}{3}A$<br\/>$I_{3}^{{}}=\\dfrac{U_{AB}^{{}}}{R_{3}^{{}}}=\\dfrac{4}{6}=\\dfrac{2}{3}A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2787},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 15\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a l\u00e0 2A; $R_{2}^{{}}$ = 15\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a b\u1eb1ng 1,5A m\u1eafc n\u1ed1i ti\u1ebfp. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a c\u00f3 th\u1ec3 \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u m\u1ea1ch \u0111\u00f3 \u0111\u1ec3 khi ho\u1ea1t \u0111\u1ed9ng kh\u00f4ng c\u00f3 \u0111i\u1ec7n tr\u1edf n\u00e0o b\u1ecb h\u1ecfng.","select":["A. 52,5V","B. 55,5V","C. 25,5V","D. 25V"],"hint":"","explain":"<span class='basic_left'>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a c\u00f3 th\u1ec3 \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u m\u1ea1ch \u0111\u1ec3 khi ho\u1ea1t \u0111\u1ed9ng kh\u00f4ng c\u00f3 \u0111i\u1ec7n tr\u1edf n\u00e0o b\u1ecb h\u1ecfng:<br\/> $U_{\\max }^{{}}=U_{\\max 1}^{{}}+U_{\\max 2}^{{}}=15\\times 2+15\\times 1,5=52,5V$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2788},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"M\u1ed9t m\u1ea1ch \u0111i\u1ec7n g\u1ed3m ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$; $R_{2}^{{}}$ v\u00e0 $R_{3}^{{}}$ m\u1eafc song song. Khi d\u00f2ng \u0111i\u1ec7n qua c\u00e1c \u0111i\u1ec7n tr\u1edf b\u1eb1ng nhau ta c\u00f3 th\u1ec3 k\u1ebft lu\u1eadn c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$; $R_{2}^{{}}$; $R_{3}^{{}}$ nh\u01b0 th\u1ebf n\u00e0o?","select":["A. $R_{1}^{{}}$ l\u1edbn nh\u1ea5t.","B. $R_{2}^{{}}$ l\u1edbn nh\u1ea5t.","C. $R_{3}^{{}}$ l\u1edbn nh\u1ea5t.","D. Ba \u0111i\u1ec7n tr\u1edf b\u1eb1ng nhau."],"hint":"","explain":"<span class='basic_left'>Ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$; $R_{2}^{{}}$ v\u00e0 $R_{3}^{{}}$ m\u1eafc song song. Khi d\u00f2ng \u0111i\u1ec7n qua c\u00e1c \u0111i\u1ec7n tr\u1edf b\u1eb1ng nhau ta c\u00f3 th\u1ec3 k\u1ebft lu\u1eadn c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$; $R_{2}^{{}}$; $R_{3}^{{}}$ b\u1eb1ng nhau v\u00ec $I=\\dfrac{U}{R}$, m\u1eafc song song n\u00ean U l\u00e0 b\u1eb1ng nhau, n\u1ebfu I b\u1eb1ng nhau th\u00ec R ph\u1ea3i b\u1eb1ng nhau.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2789},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"\u0110\u1ed1i v\u1edbi m\u1ea1ch \u0111i\u1ec7n g\u1ed3m c\u00e1c \u0111i\u1ec7n tr\u1edf m\u1eafc song song th\u00ec:","select":["A. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua c\u00e1c \u0111i\u1ec7n tr\u1edf l\u00e0 nh\u01b0 nhau.","B. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf b\u1eb1ng nhau.","C. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u m\u1ea1ch b\u1eb1ng t\u1ed5ng hi\u1ec7u \u0111i\u1ec7n th\u1ebf \u1edf hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf.","D. \u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a m\u1ea1ch b\u1eb1ng t\u1ed5ng c\u00e1c \u0111i\u1ec7n tr\u1edf th\u00e0nh ph\u1ea7n."],"hint":"","explain":"<span class='basic_left'>\u0110\u1ed1i v\u1edbi m\u1ea1ch \u0111i\u1ec7n g\u1ed3m c\u00e1c \u0111i\u1ec7n tr\u1edf m\u1eafc song song th\u00ec hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf b\u1eb1ng nhau.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2790}],"lesson":{"save":0,"level":2}}