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{"segment":[{"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":"M\u1ed9t \u0111i\u1ec7n tr\u1edf 10\u03a9 \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 12V. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111i\u1ec7n tr\u1edf \u0111\u00f3?","select":["A. 12A","B. 1,2A","C. 10A","D. 1,12A"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111i\u1ec7n tr\u1edf \u0111\u00f3: $I=\\dfrac{U}{R}=\\dfrac{12}{10}=1,2A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2751},{"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 hai \u0111i\u1ec7n tr\u1edf, $R_{1}^{{}}$=20\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 t\u1ed1i \u0111a 2A v\u00e0 $R_{2}^{{}}$=40\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 t\u1ed1i \u0111a 1,5A. 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 $R_{1}^{{}}$ n\u1ed1i ti\u1ebfp v\u1edbi $R_{2}^{{}}$ l\u00e0:","select":["A. 210V","B. 120V","C. 90V","D. 100V"],"hint":"","explain":"<span class='basic_left'>V\u00ec Khi $R_{1}^{{}}$, $R_{2}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp th\u00ec d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua hai \u0111i\u1ec7n tr\u1edf c\u00f3 c\u00f9ng c\u01b0\u1eddng \u0111\u1ed9.<br\/>Do \u0111\u00f3 m\u1ea1ch n\u00e0y ch\u1ec9 ch\u1ecbu \u0111\u01b0\u1ee3c c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a l\u00e0 1,5 A.<br\/>V\u1eady hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a l\u00e0:<br\/>$U=I\\times (R_{1}^{{}}+R_{2}^{{}})=1,5\\times (20+40)=90V$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2752},{"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 $R_{1}^{{}}$=5\u03a9, $R_{2}^{{}}$=10\u03a9, $R_{3}^{{}}$=15\u03a9 \u0111\u01b0\u1ee3c m\u1eafc n\u1ed1i ti\u1ebfp 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. 50\u03a9.","B. 20\u03a9.","C. 40\u03a9.","D. 30\u03a9."],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch l\u00e0:<br\/>$R_{t\\text{d}}^{{}}=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}=5+10+15=30\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2753},{"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":"Ph\u00e1t bi\u1ec3u n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng \u0111\u00fang \u0111\u1ed1i v\u1edbi \u0111o\u1ea1n m\u1ea1ch g\u1ed3m c\u00e1c \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp?","select":["A. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0 nh\u01b0 nhau t\u1ea1i m\u1ecdi v\u1ecb tr\u00ed c\u1ee7a \u0111o\u1ea1n m\u1ea1ch.","B. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch b\u1eb1ng t\u1ed5ng c\u00e1c hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf m\u1eafc trong \u0111o\u1ea1n m\u1ea1ch.","C. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch b\u1eb1ng hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf m\u1eafc trong \u0111o\u1ea1n m\u1ea1ch.","D. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf m\u1eafc trong \u0111o\u1ea1n m\u1ea1ch t\u1ec9 l\u1ec7 thu\u1eadn v\u1edbi \u0111i\u1ec7n tr\u1edf \u0111\u00f3."],"hint":"","explain":"<span class='basic_left'>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch b\u1eb1ng t\u1ed5ng \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf m\u1eafc trong \u0111o\u1ea1n m\u1ea1ch. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":2754},{"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":"\u0110o\u1ea1n m\u1ea1ch g\u1ed3m c\u00e1c \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp l\u00e0 \u0111o\u1ea1n m\u1ea1ch kh\u00f4ng c\u00f3 \u0111\u1eb7c \u0111i\u1ec3m n\u00e0o d\u01b0\u1edbi \u0111\u00e2y? ","select":["A. \u0110o\u1ea1n m\u1ea1ch c\u00f3 nh\u1eefng \u0111i\u1ec3m n\u1ed1i chung c\u1ee7a nhi\u1ec1u \u0111i\u1ec7n tr\u1edf.","B. \u0110o\u1ea1n m\u1ea1ch c\u00f3 nh\u1eefng \u0111i\u1ec3m n\u1ed1i chung ch\u1ec9 c\u1ee7a hai \u0111i\u1ec7n tr\u1edf.","C. D\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua c\u00e1c \u0111i\u1ec7n tr\u1edf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch c\u00f3 c\u00f9ng c\u01b0\u1eddng \u0111\u1ed9.","D. \u0110o\u1ea1n m\u1ea1ch c\u00f3 nh\u1eefng \u0111i\u1ec7n tr\u1edf m\u1eafc li\u00ean ti\u1ebfp v\u1edbi nhau v\u00e0 kh\u00f4ng c\u00f3 m\u1ea1ch r\u1ebd."],"hint":"","explain":"<span class='basic_left'>\u0110o\u1ea1n m\u1ea1ch c\u00f3 nh\u1eefng \u0111i\u1ec3m n\u1ed1i chung c\u1ee7a nhi\u1ec1u \u0111i\u1ec7n tr\u1edf l\u00e0 \u0111\u1eb7c \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n m\u1ea1ch song song. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":1}]}],"id_ques":2755},{"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 $R_{1}^{{}}$= 5\u2126, $R_{2}^{{}}$= 10\u2126 v\u00e0 ampe k\u1ebf ch\u1ec9 0,2A \u0111\u01b0\u1ee3c m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi nhau v\u00e0o hai \u0111i\u1ec3m A, B.T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB.","select":["A. 1V","B. 2V","C. 3V","D. 4V"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$U_{1}^{{}}=I\\times R_{1}^{{}}=0,2\\times 5=1V;$<br\/>$U_{2}^{{}}=I\\times R_{2}^{{}}=0,2\\times 10=2V$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB:<br\/>$U_{AB}^{{}}=U_{1}^{{}}+U_{2}^{{}}=1+2=3V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2756},{"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":"Ba \u0111i\u1ec7n tr\u1edf c\u00f3 c\u00e1c gi\u00e1 tr\u1ecb l\u00e0 10\u2126, 20\u2126, 30\u2126. C\u00f3 bao nhi\u00eau c\u00e1ch m\u1eafc c\u00e1c \u0111i\u1ec7n tr\u1edf n\u00e0y v\u00e0o m\u1ea1ch c\u00f3 hi\u1ec7u \u0111i\u1ec7n th\u1ebf 12V \u0111\u1ec3 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 0,4A","select":["A. Ch\u1ec9 c\u00f3 1 c\u00e1ch m\u1eafc.","B. C\u00f3 2 c\u00e1ch m\u1eafc.","C. C\u00f3 3 c\u00e1ch m\u1eafc.","D. Kh\u00f4ng th\u1ec3 m\u1eafc \u0111\u01b0\u1ee3c."],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch l\u00e0:<br\/> $R_{td}^{{}}=\\dfrac{U}{I}=\\dfrac{12}{0,4}=30\\Omega$<br\/>C\u00f3 3 c\u00e1ch m\u1eafc c\u00e1c \u0111i\u1ec7n tr\u1edf \u0111\u00f3 v\u00e0o m\u1ea1ch:<br\/>C\u00e1ch th\u1ee9 nh\u1ea5t l\u00e0 ch\u1ec9 m\u1eafc \u0111i\u1ec7n tr\u1edf R = 30\u2126 trong \u0111o\u1ea1n m\u1ea1ch;<br\/>C\u00e1ch th\u1ee9 hai l\u00e0 m\u1eafc hai \u0111i\u1ec7n tr\u1edf R = 10\u2126 v\u00e0 R = 20\u2126 n\u1ed1i ti\u1ebfp nhau trong \u0111o\u1ea1n m\u1ea1ch.<br\/>C\u00e1ch th\u1ee9 ba l\u00e0 m\u1eafc ba \u0111i\u1ec7n tr\u1edf R = 10\u2126 n\u1ed1i ti\u1ebfp nhau.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2757},{"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 g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$v\u00e0 $R_{2}^{{}}= $1,5$R_{1}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi nhau. Cho d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111o\u1ea1n m\u1ea1ch n\u00e0y th\u00ec th\u1ea5y hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$l\u00e0 3V. H\u1ecfi hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch l\u00e0 bao nhi\u00eau ? ","select":["A. 1,5V.","B. 3V.","C. 4,5V.","D. 7,5V."],"hint":"","explain":"<span class='basic_left'>V\u00ec hai \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp:<br\/>$\\dfrac{U_{1}^{{}}}{U_{2}^{{}}}=\\dfrac{I\\times R_{1}^{{}}}{I\\times R_{2}^{{}}}=\\dfrac{R_{1}^{{}}}{1,5R_{1}^{{}}}$<br\/>$\\Rightarrow U_{2}^{{}}=1,5U_{1}^{{}}=1,5\\times 3=4,5V$<br\/>$\\Rightarrow U=U_{1}^{{}}+U_{2}^{{}}=3+4,5=7,5V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2758},{"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":"\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf $U_{AB}^{{}}$v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$v\u00e0 $R_{2}^{{}}$m\u1eafc n\u1ed1i ti\u1ebfp. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u1ee9ng l\u00e0 $U_{1}^{{}}$, $U_{2}^{{}}$. H\u1ec7 th\u1ee9c n\u00e0o d\u01b0\u1edbi \u0111\u00e2y l\u00e0 kh\u00f4ng \u0111\u00fang ?","select":["A. $R_{AB}^{{}}=R_{1}^{{}}+R_{2}^{{}}$","B. $I_{AB}^{{}}=I_{1}^{{}} = I_{2}^{{}}$","C. $\\dfrac{U_{1}^{{}}}{U_{2}^{{}}}=\\dfrac{R_{2}^{{}}}{R_{1}^{{}}}$","D. $U_{AB}^{{}}=U_{1}^{{}}+U_{2}^{{}}$."],"hint":"","explain":"<span class='basic_left'>$\\dfrac{U_{1}^{{}}}{U_{2}^{{}}}=\\dfrac{R_{2}^{{}}}{R_{1}^{{}}}$ \u0110\u00e2y l\u00e0 h\u1ec7 th\u1ee9c kh\u00f4ng li\u00ean quan \u0111\u1ebfn \u0111o\u1ea1n m\u1ea1ch n\u1ed1i ti\u1ebfp.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2759},{"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/bai4/lv2/img\/h10.jpg'\/><\/center>Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ v\u00e0 ampe k\u1ebf \u0111\u01b0\u1ee3c m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi nhau v\u00e0o hai \u0111i\u1ec3m A v\u00e0 B nh\u01b0 h\u00ecnh v\u1ebd. Cho $R_{1}^{{}}$ = 5\u03a9, $R_{2}^{{}}$ = 10\u03a9, ampe k\u1ebf ch\u1ec9 0,2A. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB.","select":["A. 1V.","B. 2V.","C. 3V.","D. 4V."],"hint":"","explain":"<span class='basic_left'>$R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=5+10=15\\Omega $<br\/>$\\Rightarrow U_{AB}^{{}}=I\\times R_{td}^{{}}=0,2\\times 15=3V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2760},{"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/bai4/lv2/img\/h11.jpg'\/><\/center>C\u00f3 m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 10\u03a9, $R_{2}^{{}}$ = 20\u03a9, hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch AB b\u1eb1ng 12V. S\u1ed1 ch\u1ec9 c\u1ee7a v\u00f4n k\u1ebf v\u00e0 ampe k\u1ebf l\u00e0 bao nhi\u00eau?","select":["A. U=1V; I=0,1A","B. U=2V; I=0,2A","C. U=3V; I=0,3A","D. U=4V; I=0,4A"],"hint":"","explain":"<span class='basic_left'>Do hai \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp:<br\/>$R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=10+20=30\\Omega $<br\/>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf l\u00e0: $I=\\dfrac{U_{AB}^{{}}}{R_{td}^{{}}}=\\dfrac{12}{30}=0,4A$<br\/>S\u1ed1 ch\u1ec9 c\u1ee7a v\u00f4n k\u1ebf l\u00e0: $U=I\\times R_{1}^{{}}=0,4\\times 10=4V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2761},{"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/bai4/lv2/img\/h12.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 c\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 5\u03a9, $R_{2}^{{}}$ = 15\u03a9. V\u00f4n k\u1ebf ch\u1ec9 3V. S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf l\u00e0 bao nhi\u00eau? T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u AB c\u1ee7a \u0111o\u1ea1n m\u1ea1ch.","select":["A. U=4V; I=0,2A","B. U=2V; I=0,1A","C. U=3V; I=0,3A","D. U=4V; I=0,4A"],"hint":"","explain":"<span class='basic_left'>S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf l\u00e0:<br\/> $I=\\dfrac{U_{2}^{{}}}{R_{2}^{{}}}=\\dfrac{3}{15}=0,2A$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u AB c\u1ee7a \u0111o\u1ea1n m\u1ea1ch l\u00e0:<br\/> $U_{AB}^{{}}=I\\times R_{td}^{{}}=I\\times (R_{1}^{{}}+R_{2}^{{}})=0,2\\times (5+15)=4V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2762},{"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\u1ec7n tr\u1edf $R_{1}^{{}}$ = 5\u03a9, $R_{2}^{{}}$= 10\u03a9, $R_{3}^{{}}$ = 15\u03a9 \u0111\u01b0\u1ee3c m\u1eafc n\u1ed1i ti\u1ebfp nhau v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 12V. T\u00ednh \u0111i\u1ec7n tr\u1edf tr\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf.","select":["A. $R_{td}^{{}}$=20\u03a9; $U_{1}^{{}}$ =1V; $U_{2}^{{}}$ =2V; $U_{3}^{{}}$=3V.","B. $R_{td}^{{}}$=30\u03a9; $U_{1}^{{}}$ =2V; $U_{2}^{{}}$ =4V; $U_{3}^{{}}$=6V.","C. $R_{td}^{{}}$=40\u03a9; $U_{1}^{{}}$ =3V; $U_{2}^{{}}$ =6V; $U_{3}^{{}}$=9V.","D. $R_{td}^{{}}$=50\u03a9; $U_{1}^{{}}$ =4V; $U_{2}^{{}}$ =8V; $U_{3}^{{}}$=12V."],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch l\u00e0:<br\/>$R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}=5+10+15=30\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua d\u00e2y d\u1eabn: <br\/>$I=\\dfrac{U}{R}=\\dfrac{12}{30}=0,4A$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u m\u1ed7i \u0111i\u1ec7n tr\u1edf:<br\/>$U_{1}^{{}}=I\\times R_{1}^{{}}=0,4\\times 5=2V$<br\/>$U_{2}^{{}}=I\\times R_{2}^{{}}=0,4\\times 10=4V$<br\/>$U_{3}^{{}}=I\\times R_{3}^{{}}=0,4\\times 15=6V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2763},{"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":"\u0110\u1eb7t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 12V v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 40\u03a9 v\u00e0 $R_{2}^{{}}$ = 80\u03a9 m\u1eafc n\u1ed1i ti\u1ebfp. H\u1ecfi c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ea1ch n\u00e0y l\u00e0 bao nhi\u00eau?","select":["A. 0,1A","B. 0,15A","C. 0,45A","D. 0,3A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng:<br\/>$R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=40+80=120\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ea1ch:<br\/>$I=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{12}{120}=0,1A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2764},{"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/bai4/lv2/img\/h15.jpg'\/><\/center>\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U v\u00e0o hai \u0111\u1ea7u m\u1ed9t \u0111o\u1ea1n m\u1ea1ch c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 tr\u00ean h\u00ecnh v\u1ebd, trong \u0111\u00f3 c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 3\u03a9, $R_{2}^{{}}$ = 6\u03a9. H\u1ecfi s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf khi c\u00f4ng t\u1eafc K \u0111\u00f3ng l\u1edbn h\u01a1n hay nh\u1ecf h\u01a1n bao nhi\u00eau l\u1ea7n so v\u1edbi khi c\u00f4ng t\u1eafc K m\u1edf?","select":["A. Nh\u1ecf h\u01a1n 2 l\u1ea7n","B. L\u1edbn h\u01a1n 2 l\u1ea7n","C. Nh\u1ecf h\u01a1n 3 l\u1ea7n","D. L\u1edbn h\u01a1n 3 l\u1ea7n"],"hint":"","explain":"<span class='basic_left'>Khi c\u00f4ng t\u1eafc K m\u1edf m\u1ea1ch g\u1ed3m $R_{1}^{{}}$nt $R_{2}^{{}}$ nt ampe k\u1ebf n\u00ean \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a m\u1ea1ch l\u00e0: $R=R_{1}^{{}}+R_{2}^{{}}=9\\Omega $ <br\/>n\u00ean s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf l\u00e0:<br\/>$I_{1}^{{}}=\\dfrac{U}{R}=\\dfrac{U}{9}$<br\/>Khi c\u00f4ng t\u1eafc K \u0111\u00f3ng th\u00ec $R_{2}^{{}}$ b\u1ecb \u0111\u1ea5u t\u1eaft, m\u1ea1ch ch\u1ec9 c\u00f2n ($R_{1}^{{}}$ nt Ampe k\u1ebf) <br\/> \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a m\u1ea1ch l\u00e0 $R_{1}^{{}}$ = 3 <br\/>n\u00ean s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf l\u00e0: $I_{2}^{{}}=\\dfrac{U}{R_{1}^{{}}}=\\dfrac{U}{3}$<br\/>Ta c\u00f3:<br\/>$\\dfrac{I_{2}^{{}}}{I_{1}^{{}}}=\\dfrac{\\dfrac{U}{3}}{\\dfrac{U}{9}}=3$ n\u00ean s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf khi c\u00f4ng t\u1eafc K \u0111\u00f3ng l\u1edbn h\u01a1n 3 l\u1ea7n so v\u1edbi khi c\u00f4ng t\u1eafc K m\u1edf.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2765},{"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":"\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 6V v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m ba \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=3\u03a9, $R_{2}^{{}}$ = 5\u03a9, $R_{3}^{{}}$ = 7\u03a9 m\u1eafc n\u1ed1i ti\u1ebfp. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ed7i \u0111i\u1ec7n tr\u1edf c\u1ee7a \u0111o\u1ea1n m\u1ea1ch.","select":["A. 0,1A","B. 0,2A","C. 0,4A","D. 0,6A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng:<br\/>$R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}=3+5+7=15\\Omega $<br\/> Do ba \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp n\u00ean c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua m\u1ed7i \u0111i\u1ec7n tr\u1edf:<br\/>$I=I_{1}^{{}}=I_{2}^{{}}=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{6}{15}=0,4A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2766},{"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/bai4/lv2/img\/h17.jpg'\/><\/center>\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 tr\u00ean h\u00ecnh v\u1ebd trong \u0111\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 4\u03a9 , $R_{2}^{{}}$ = 5\u03a9. Cho bi\u1ebft s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf khi c\u00f4ng t\u1eafc K m\u1edf v\u00e0 khi K \u0111\u00f3ng h\u01a1n k\u00e9m nhau 3 l\u1ea7n. Cho bi\u1ebft U = 5,4V. S\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf khi c\u00f4ng t\u1eafc K m\u1edf l\u00e0 bao nhi\u00eau?","select":["A. 0,1A","B. 0,2A","C. 0,4A","D. 0,6A"],"hint":"","explain":"<span class='basic_left'>$I_{d}^{{}}=3I_{m}^{{}}\\Leftrightarrow \\dfrac{U_{1}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{3U_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}}$<br\/>$\\Leftrightarrow \\dfrac{1}{4+5}=\\dfrac{3}{4+5+R_{3}^{{}}}$<br\/>$\\Rightarrow R_{3}^{{}}=18\\Omega $<br\/>V\u1eady s\u1ed1 ch\u1ec9 c\u1ee7a ampe k\u1ebf khi c\u00f4ng t\u1eafc K m\u1edf:<br\/>$I=\\dfrac{U}{R}=\\dfrac{5,4}{27}=0,2A$<br\/><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2767},{"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/bai4/lv2/img\/mach1.png'\/><\/center>\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U v\u00e0o hai \u0111\u1ea7u m\u1ed9t \u0111o\u1ea1n m\u1ea1ch c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 tr\u00ean h\u00ecnh v\u1ebd. Khi \u0111\u00f3ng c\u00f4ng t\u1eafc K v\u00e0o v\u1ecb tr\u00ed 1 th\u00ec ampe k\u1ebf c\u00f3 s\u1ed1 ch\u1ec9 $I_{1}^{{}}=I;$khi chuy\u1ec3n c\u00f4ng t\u1eafc n\u00e0y sang v\u1ecb tr\u00ed s\u1ed1 2 th\u00ec ampe k\u1ebf c\u00f3 s\u1ed1 ch\u1ec9 l\u00e0 $I_{2}^{{}}=\\dfrac{I}{3};$c\u00f2n khi chuy\u1ec3n K sang v\u1ecb tr\u00ed 3 th\u00ec ampe k\u1ebf c\u00f3 s\u1ed1 ch\u1ec9 $I_{3}^{{}}=\\dfrac{I}{8}_{{}}^{{}};$ Cho bi\u1ebft $R_{1}^{{}}$ = 3\u03a9, h\u00e3y t\u00ednh $R_{2}^{{}}$ v\u00e0 $R_{3}^{{}}$.","select":["A. $R_{2}^{{}}$ = 5\u03a9, $R_{3}^{{}}$ = 12\u03a9","B. $R_{2}^{{}}$ = 6\u03a9, $R_{3}^{{}}$ = 15\u03a9","C. $R_{2}^{{}}$ = 7\u03a9, $R_{3}^{{}}$ = 18\u03a9","D. $R_{2}^{{}}$ = 9\u03a9, $R_{3}^{{}}$ = 21\u03a9"],"hint":"","explain":"<span class='basic_left'>Khi K \u0111\u00f3ng \u1edf v\u1ecb tr\u00ed 1: $I_{1}^{{}}=I;$ $R_{1}^{{}}=30\\Omega \\Rightarrow U=I_{1}^{{}}\\times R_{1}^{{}}=3I$ (1)<br\/>Khi K \u0111\u00f3ng \u1edf v\u1ecb tr\u00ed 2: $I_{2}^{{}}=\\dfrac{I}{3};$$R=R_{1}^{{}}+R_{2}^{{}}=3+R_{2}^{{}}\\Rightarrow U=I_{2}^{{}}\\times (3+R_{2}^{{}})=\\dfrac{I}{3}\\times (3+R_{2}^{{}})$ (2)<br\/>Khi K \u0111\u00f3ng \u1edf v\u1ecb tr\u00ed 3: $I_{3}^{{}}=\\dfrac{I}{8}_{{}}^{{}};$$R=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}=3+R_{2}^{{}}+R_{3}^{{}}$<br\/>$\\Rightarrow U=I_{3}^{{}}\\times R=\\dfrac{I}{8}\\times (3+R_{2}^{{}}+R_{3}^{{}})$ (3)<br\/>Thay (1) v\u00e0o (2) ta c\u00f3: $3I=\\dfrac{I}{3}\\times (3+R_{2}^{{}})\\Rightarrow R_{2}^{{}}=6\\Omega $<br\/>Thay (1) v\u00e0 $R_{2}^{{}}$ v\u00e0o (3) ta c\u00f3: $3I=\\dfrac{I}{8}\\times (3+6+R_{3}^{{}})\\Rightarrow R_{3}^{{}}=15\\Omega $.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2768},{"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 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 100\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a l\u00e0 0,6A v\u00e0 \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$ = 50\u03a9 ch\u1ecbu \u0111\u01b0\u1ee3c c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a l\u00e0 0,4A. C\u00f3 th\u1ec3 m\u1eafc n\u1ed1i ti\u1ebfp hai \u0111i\u1ec7n tr\u1edf tr\u00ean v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a l\u00e0:","select":["A. U = 80V","B. U = 60V","C. U = 90V","D. U = 30V"],"hint":"","explain":"<span class='basic_left'>V\u00ec Khi $R_{1}^{{}}$, $R_{2}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp th\u00ec d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua hai \u0111i\u1ec7n tr\u1edf c\u00f3 c\u00f9ng c\u01b0\u1eddng \u0111\u1ed9.<br\/>Do \u0111\u00f3 m\u1ea1ch n\u00e0y ch\u1ec9 ch\u1ecbu \u0111\u01b0\u1ee3c c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n t\u1ed1i \u0111a l\u00e0 0,4 A.<br\/> V\u1eady hi\u1ec7u \u0111i\u1ec7n th\u1ebf t\u1ed1i \u0111a l\u00e0: $U=I\\times (R_{1}^{{}}+R_{2}^{{}})=0,4\\times (100+50)=60V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2769},{"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":"M\u1ed9t m\u1ea1ch \u0111i\u1ec7n g\u1ed3m $R_{1}^{{}}$ n\u1ed1i ti\u1ebfp $R_{2}^{{}}$. \u0110i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 4\u03a9, $R_{2}^{{}}$ = 6\u03a9. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u m\u1ea1ch l\u00e0 U = 12V. Hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u $R_{2}^{{}}$ l\u00e0:","select":["A. 4V","B. 4,8V","C. 7,2V","D. 13V"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n: <br\/>$I=I_{1}^{{}}=I_{2}^{{}}=\\dfrac{U}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{12}{4+6}=1,2A$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u $R_{2}^{{}}$ l\u00e0:<br\/> $U_{2}^{{}}=I\\times R_{2}^{{}}=1,2\\times 6=7,2V$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2770}],"lesson":{"save":0,"level":2}}

Điểm của bạn.

Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý