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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":[[3]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/vatly/bai6/lv2/img\/h1.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n AB c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=3r; $R_{2}^{{}}$=r; $R_{3}^{{}}$=6r; \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch n\u00e0y c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o d\u01b0\u1edbi \u0111\u00e2y? ","select":["A. 0,75r","B. 3r","C. 2,1r","D. 10r"],"hint":"","explain":"<span class='basic_left'>Ta th\u1ea5y $R_{1}^{{}}$ \/\/ ($R_{2}^{{}}$ nt $R_{3}^{{}}$ )<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: $R_{23}^{{}}=R_{2}^{{}}+R_{3}^{{}}=r+6r=7r$<br\/>$\\Rightarrow R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{23}^{{}}}{R_{1}^{{}}+R_{23}^{{}}}=\\dfrac{3r\\times 7r}{3r+7r}=2,1r$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2911},{"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/bai6/lv2/img\/h2.jpg'\/><\/center>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 tr\u00ean h\u00ecnh v\u1ebd l\u00e0 $R_{AB}^{{}}$=10\u2126, trong \u0111\u00f3 c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=7\u2126; $R_{2}^{{}}$=12\u2126. H\u1ecfi \u0111i\u1ec7n tr\u1edf $R_{x}^{{}}$ c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o d\u01b0\u1edbi \u0111\u00e2y? ","select":["A. 9\u2126 ","B. 5\u2126 ","C. 15\u2126","D. 4\u2126"],"hint":"","explain":"<span class='basic_left'>Ta th\u1ea5y $R_{1}^{{}}$nt($R_{2}^{{}}$\/\/$R_{x}^{{}}$)<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch:<br\/> $R_{AB}^{{}}=R_{1}^{{}}+\\dfrac{R_{2}^{{}}\\times R_{x}^{{}}}{R_{2}^{{}}+R_{x}^{{}}}=10\\Omega \\Leftrightarrow 10=7+\\dfrac{12\\times R_{x}^{{}}}{12+R_{x}^{{}}}\\Leftrightarrow R_{x}^{{}}=4\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2912},{"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":"\u0110i\u1ec7n tr\u1edf $R_{1}^{{}}$=6\u2126; $R_{2}^{{}}$=9\u2126; $R_{3}^{{}}$=15\u2126 ch\u1ecbu \u0111\u01b0\u1ee3c d\u00f2ng \u0111i\u1ec7n c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 l\u1edbn nh\u1ea5t t\u01b0\u01a1ng \u1ee9ng l\u00e0 $I_{1}^{{}}$=5A, $I_{2}^{{}}$=2A, $I_{3}^{{}}$=3A. H\u1ecfi c\u00f3 th\u1ec3 \u0111\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf l\u1edbn nh\u1ea5t bao nhi\u00eau v\u00e0o ha \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch g\u1ed3m 3 \u0111i\u1ec7n tr\u1edf m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi nhau?","select":["A. 45V ","B. 60V ","C. 93V ","D. 150V"],"hint":"","explain":"<span class='basic_left'>Do ba \u0111i\u1ec7n tr\u1edf n\u00e0y m\u1eafc n\u1ed1i ti\u1ebfp n\u00ean ta c\u00f3: $I=I_{1}^{{}}=I_{2}^{{}}=I_{3}^{{}}=2A$<br\/>(l\u1ea5y gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t, n\u1ebfu l\u1ea5y gi\u00e1 tr\u1ecb kh\u00e1c l\u1edbn h\u01a1n th\u00ec \u0111i\u1ec7n tr\u1edf b\u1ecb h\u1ecfng).<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf to\u00e0n m\u1ea1ch U theo \u0111\u1ecbnh lu\u1eadt \u00d4m.<br\/>$U=\\text{I}\\times \\text{R}=I\\times (R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}})=60V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2913},{"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":"Khi m\u1eafc n\u1ed1i ti\u1ebfp hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 1,2V th\u00ec d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua ch\u00fang c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 I=0,12A. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch n\u1ed1i ti\u1ebfp n\u00e0y. N\u1ebfu m\u1eafc song song hai \u0111i\u1ec7n tr\u1edf n\u00f3i tr\u00ean v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf th\u00ec d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 $I_{1}^{{}}$ g\u1ea5p 1,5 l\u1ea7n c\u01b0\u1eddng \u0111\u1ed9 $I_{2}^{{}}$ c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$. H\u00e3y t\u00ednh \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$v\u00e0 $R_{2}^{{}}$.","select":["A. $R_{td}^{{}}$=10\u2126; $R_{1}^{{}}$=4\u2126; $R_{2}^{{}}$=6\u2126","B. $R_{td}^{{}}$=10\u2126; $R_{1}^{{}}$=6\u2126; $R_{2}^{{}}$=4\u2126 ","C. $R_{td}^{{}}$=2,4\u2126; $R_{1}^{{}}$=4\u2126; $R_{2}^{{}}$=6\u2126 ","D. $R_{td}^{{}}$=2,4\u2126; $R_{1}^{{}}$ =6\u2126; $R_{2}^{{}}$=4\u2126"],"hint":"","explain":"<span class='basic_left'>Theo \u0111\u1ecbnh lu\u1eadt \u00d4m: $R_{td}^{{}}=\\dfrac{U}{I}=R_{1}^{{}}+R_{2}^{{}}=10\\Omega $ (1)<br\/>Ta c\u00f3 $U_{1}^{{}}=U_{2}^{{}}\\Leftrightarrow I_{1}^{{}}\\times R_{1}^{{}}=I_{2}^{{}}\\times R_{2}^{{}}\\Rightarrow R_{1}^{{}}=\\dfrac{I_{2}^{{}}R_{2}^{{}}}{I_{1}^{{}}}=\\dfrac{R_{2}^{{}}}{1,5}$ (2)<br\/>T\u1eeb (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c: $R_{1}^{{}}+R_{2}^{{}}=10\\Leftrightarrow \\dfrac{R_{2}^{{}}}{1,5}+R_{2}^{{}}=10\\Leftrightarrow R_{2}^{{}}\\times (1+\\dfrac{1}{1,5})=10\\Leftrightarrow R_{2}^{{}}=6\\Omega $<br\/>$\\Rightarrow R_{1}^{{}}=4\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2914},{"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/bai6/lv2/img\/h5.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=14\u2126; $R_{2}^{{}}$=8\u2126; $R_{3}^{{}}$=24\u2126; d\u00f2ng \u0111i\u1ec7n \u0111i qua $R_{1}^{{}}$ c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 l\u00e0 $I_{1}^{{}}$=0,4A. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n $I_{2}^{{}}$, $I_{3}^{{}}$ t\u01b0\u01a1ng \u1ee9ng \u0111i qua c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{2}^{{}}$ v\u00e0 $R_{3}^{{}}$.","select":["A. $I_{2}^{{}}$=0,1A, $I_{3}^{{}}$=0,3A","B. $I_{2}^{{}}$= 3A, $I_{3}^{{}}$= 1A ","C. $I_{2}^{{}}$=0,1A, $I_{3}^{{}}$=0,1A","D. $I_{2}^{{}}$=0,3A, $I_{3}^{{}}$=0,1A"],"hint":"","explain":"<span class='basic_left'>Ta th\u1ea5y $I_{1}^{{}}=I_{23}^{{}}=0,4A$<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: $R_{AB}^{{}}=R_{1}^{{}}+\\dfrac{R_{2}^{{}}\\times R_{3}^{{}}}{R_{2}^{{}}+R_{3}^{{}}}=14+\\dfrac{8\\times 24}{8+24}=20\\Omega $<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf c\u1ee7a m\u1ea1ch: $U=\\text{I}\\times \\text{R}_{AB}^{{}}=0,4\\times 20=8V$<br\/>$U_{1}^{{}}=I_{1}^{{}}\\times R_{1}^{{}}=0,4\\times 14=5,6V$<br\/>$U_{23}^{{}}=U-U_{1}^{{}}=8-5,6=2,4V=U_{2}^{{}}=U_{3}^{{}}$<br\/>$I_{2}^{{}}=\\dfrac{U_{2}^{{}}}{R_{2}^{{}}}=0,3A$<br\/>$I_{3}^{{}}=\\dfrac{U_{3}^{{}}}{R_{3}^{{}}}=0,1A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2915},{"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/bai6/lv2/img\/h6.jpg'\/><\/center>Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ \u0111\u01b0\u1ee3c m\u1eafc theo hai c\u00e1ch v\u00e0o hai \u0111i\u1ec3m M, N theo s\u01a1 \u0111\u1ed3 , trong \u0111\u00f3 hi\u1ec7u \u0111i\u1ec7n th\u1ebf U=6V. Trong c\u00e1ch m\u1eafc th\u1ee9 nh\u1ea5t, ampe k\u1ebf ch\u1ec9 0,4A. Trong c\u00e1ch m\u1eafc th\u1ee9 hai, ampe k\u1ebf ch\u1ec9 1,8A. T\u00ednh \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$. ","select":["A. $R_{1}^{{}}$ = 5 \u03a9; $R_{2}^{{}}$ = 5 \u03a9","B. $R_{1}^{{}}$ = 5 \u03a9; $R_{2}^{{}}$ = 10 \u03a9","C. $R_{1}^{{}}$ = 3 \u03a9; $R_{2}^{{}}$ = 6 \u03a9 ","D. $R_{1}^{{}}$ = 6 \u03a9; $R_{2}^{{}}$ = 4 \u03a9"],"hint":"","explain":"<span class='basic_left'>Khi $R_{1}^{{}}$ n\u1ed1i ti\u1ebfp $R_{2}^{{}}$ th\u00ec $I_{1}^{{}}$=0,4A <br\/>$R_{1}^{{}}+R_{2}^{{}}=\\dfrac{U}{I_{1}^{{}}}=\\dfrac{6}{0,4}=15\\Omega $ (1)<br\/>Khi $R_{1}^{{}}$ song song $R_{2}^{{}}$th\u00ec $I_{2}^{{}}$=1,8A : <br\/>$\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{U}{I_{2}^{{}}}=\\dfrac{6}{1,8}=\\dfrac{60}{18}\\Omega $ (2)<br\/>K\u1ebft h\u1ee3p (1) v\u00e0 (2) ta c\u00f3: $R_{1}^{{}}\\times R_{2}^{{}}=50\\Omega $ (3)<br\/>T\u1eeb (1) v\u00e0 (3) gi\u1ea3i ra ta c\u00f3: $R_{1}^{{}}=5\\Omega ;$ $R_{2}^{{}}=10\\Omega $ho\u1eb7c $R_{1}^{{}}=10\\Omega ;$$R_{2}^{{}}=5\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2916},{"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":"Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=$R_{2}^{{}}$=20\u03a9 \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hai \u0111i\u1ec3m A, B. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng $R_{td}^{{}}$ c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB khi $R_{1}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi $R_{2}^{{}}$ ","select":["A. 40\u2126","B. 30\u2126","C. 60\u2126","D. 50\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_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=20+20=40\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2917},{"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/bai6/lv2/img\/h8.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n c\u00f3 s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 c\u00f3 c\u00e1c \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$=9\u2126; $R_{2}^{{}}$=15\u2126; $R_{3}^{{}}$=10\u2126; d\u00f2ng \u0111i\u1ec7n \u0111i qua $R_{3}^{{}}$ c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 l\u00e0 $I_{3}^{{}}$=0,3A.T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf U gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch AB. ","select":["A. 6,5V","B. 2,5V","C. 7,5V","D. 5,5V"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a $R_{2}^{{}}$ v\u00e0 $R_{3}^{{}}$: $R_{23}^{{}}=\\dfrac{R_{2}^{{}}\\times R_{3}^{{}}}{R_{2}^{{}}+R_{3}^{{}}}=\\dfrac{15\\times 10}{15+10}=6\\Omega $<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u $R_{3}^{{}}$ : $U_{3}^{{}}=I_{3}^{{}}\\times R_{3}^{{}}=0,3\\times 10=3V=U_{2}^{{}}=U_{23}^{{}}$<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua $R_{2}^{{}}$: $I_{2}^{{}}=\\dfrac{U_{2}^{{}}}{R_{2}^{{}}}=\\dfrac{3}{15}=0,2A$<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua $R_{1}^{{}}$:$I=I_{1}^{{}}=I_{2}^{{}}+I_{3}^{{}}=0,3+0,2=0,5A$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch AB l\u00e0: $U_{AB}^{{}}=\\text{I}\\times \\text{R}=I\\times \\left( R_{23}^{{}}+R_{1}^{{}} \\right)=0,5\\times \\left( 6+9 \\right)=7,5V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2918},{"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":"Hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = $R_{2}^{{}}$ = 20\u03a9 \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hai \u0111i\u1ec3m A, B. \u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch AB l\u00e0 $R_{td}^{{}}$ khi $R_{1}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp v\u1edbi $R_{2}^{{}}$. Khi m\u1eafc $R_{1}^{{}}$song song v\u1edbi $R_{2}^{{}}$ th\u00ec \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng l\u00e0 $R_{td}^{'}$. T\u00ednh t\u1ef7 s\u1ed1 gi\u1eefa $R_{td}^{{}}$ v\u00e0 $R_{td}^{'}$.","select":["A. 1","B. 2","C. 3","D. 4"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng khi m\u1eafc n\u1ed1i ti\u1ebfp: $R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=20+20=40\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng khi m\u1eafc song song: $R_{td}^{'}$=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{20\\times 20}{20+20}=10\\Omega $<br\/>T\u1ec9 s\u1ed1 gi\u1eefa $R_{td}^{{}}$ v\u00e0 $R_{td}^{'}$ l\u00e0:<br\/>$\\dfrac{R_{td}^{{}}}{R_{td}^{{'}}}=\\dfrac{40}{10}=4$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2919},{"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/bai6/lv2/img\/h10.jpg'\/><\/center>C\u00e1c \u0111i\u1ec7n tr\u1edf R l\u00e0 nh\u01b0 nhau trong c\u00e1c \u0111o\u1ea1n m\u1ea1ch c\u00f3 s\u01a1 \u0111\u1ed3 trong h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y. H\u1ecfi \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch n\u00e0o l\u00e0 nh\u1ecf nh\u1ea5t?","select":["A. H\u00ecnh A","B. H\u00ecnh B","C. H\u00ecnh C","D. H\u00ecnh D"],"hint":"","explain":"<span class='basic_left'>V\u00ec khi m\u1eafc song song \u0111i\u1ec7n tr\u1edf to\u00e0n m\u1ea1ch s\u1ebd nh\u1ecf h\u01a1n m\u1ed7i \u0111i\u1ec7n tr\u1edf trong m\u1ea1ch n\u00ean h\u00ecnh D c\u00f3 \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng nh\u1ecf nh\u1ea5t.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2920},{"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/bai6/lv2/img\/h11.jpg'\/><\/center>Cho ba \u0111i\u1ec7n tr\u1edf l\u00e0 $R_{1}^{{}}$ = 6\u03a9 ; $R_{2}^{{}}$ = 12\u03a9 v\u00e0 $R_{3}^{{}}$ = 18\u03a9 \u0111\u01b0\u1ee3c m\u1eafc nh\u01b0 h\u00ecnh v\u1ebd. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng.","select":["A. 3\u2126","B. 6\u2126","C. 9\u2126","D. 12\u2126"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 $\\left( R_{1}^{{}}ntR_{2}^{{}} \\right)\/\/R_{3}^{{}}$ n\u00ean:$R_{12}^{{}}=R_{1}^{{}}+R_{2}^{{}}=6+12=18\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: $R_{td}^{{}}=\\dfrac{R_{12}^{{}}\\times R_{3}^{{}}}{R_{12}^{{}}+R_{3}^{{}}}=\\dfrac{18\\times 18}{18+18}=9\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2921},{"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/bai6/lv2/img\/h12.jpg'\/><\/center>Cho ba \u0111i\u1ec7n tr\u1edf l\u00e0 $R_{1}^{{}}$ = 6\u03a9 ; $R_{2}^{{}}$ = 12\u03a9 v\u00e0 $R_{3}^{{}}$ = 18\u03a9 \u0111\u01b0\u1ee3c m\u1eafc nh\u01b0 h\u00ecnh v\u1ebd. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng.","select":["A. 5\u2126","B. 10\u2126","C. 15\u2126","D. 20\u2126"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 $(R_{2}^{{}}ntR_{3}^{{}})\/\/R_{1}^{{}}$ n\u00ean: $R_{23}^{{}}=R_{2}^{{}}+R_{3}^{{}}=12+18=30\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: $R_{td}^{{}}=\\dfrac{R_{23}^{{}}\\times R_{1}^{{}}}{R_{23}^{{}}+R_{1}^{{}}}=\\dfrac{30\\times 6}{30+6}=5\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2922},{"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/bai6/lv2/img\/h13.jpg'\/><\/center>Cho ba \u0111i\u1ec7n tr\u1edf l\u00e0 $R_{1}^{{}}$ = 6\u03a9 ; $R_{2}^{{}}$ = 12\u03a9 v\u00e0 $R_{3}^{{}}$ = 18\u03a9 \u0111\u01b0\u1ee3c m\u1eafc nh\u01b0 h\u00ecnh v\u1ebd. T\u00ednh \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng.","select":["A. 2\u2126","B. 4\u2126","C. 8\u2126","D. 12\u2126"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3 $(R_{1}^{{}}ntR_{3}^{{}})\/\/R_{2}^{{}}$ n\u00ean: $R_{13}^{{}}=R_{1}^{{}}+R_{3}^{{}}=6+18=24\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: $R_{td}^{{}}=\\dfrac{R_{13}^{{}}\\times R_{2}^{{}}}{R_{13}^{{}}+R_{2}^{{}}}=\\dfrac{24\\times 12}{24+12}=8\\Omega $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2923},{"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 $R_{1}^{{}}$ = 3\u03a9, $R_{2}^{{}}$ = 2\u03a9 m\u1eafc n\u1ed1i ti\u1ebfp; c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch l\u00e0 0,12A. N\u1ebfu m\u1eafc song song hai \u0111i\u1ec7n tr\u1edf tr\u00ean v\u00e0o m\u1ea1ch th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0:","select":["A. 1,2A","B. 1A","C. 0,5A","D. 1,8A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng m\u1ea1ch n\u1ed1i ti\u1ebfp: $R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=3+2=5\\Omega $<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u m\u1ea1ch U l\u00e0: $U=\\text{I}\\times \\text{R}=0,12\\times 5=0,6V$<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng m\u1ea1ch song song: $R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{3\\times 2}{3+2}=1,2\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n: $I=\\dfrac{U}{R}=\\dfrac{0,6}{1,2}=0,5A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2924},{"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":"\u0110\u1eb7t m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U nh\u01b0 nhau v\u00e0o hai \u0111\u1ea7u \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$, Bi\u1ebft $R_{2}^{{}}$ = 2$R_{1}^{{}}$. N\u1ebfu hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ m\u1eafc n\u1ed1i ti\u1ebfp th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n l\u00e0 I = 0,2A. N\u1ebfu m\u1eafc hai \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ v\u00e0 $R_{2}^{{}}$ song song v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf tr\u00ean th\u00ec c\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch ch\u00ednh l\u00e0:","select":["A. 0,2A","B. 0,3A","C. 0,4A","D. 0,9A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng m\u1ea1ch n\u1ed1i ti\u1ebfp:<br\/> $R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}=3R_{1}^{{}}\\Rightarrow U=0,2\\times 3R_{1}^{{}}=0,6R_{1}^{{}}$<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng m\u1ea1ch song song:<br\/> $R_{td}^{{}}=\\dfrac{R_{1}^{{}}\\times R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}=\\dfrac{R_{1}^{{}}\\times 2R_{1}^{{}}}{R_{1}^{{}}+2R_{1}^{{}}}=\\dfrac{2}{3}R_{1}^{{}}$<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n: $I=\\dfrac{U}{R}=0,9A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2925},{"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/bai6/lv2/img\/h16.jpg'\/><\/center>Cho m\u1ea1ch \u0111i\u1ec7n nh\u01b0 h\u00ecnh v\u1ebd v\u1edbi $R_{1}^{{}}$ = 2\u03a9; $R_{2}^{{}}$ = 4\u03a9; $R_{3}^{{}}$ = 8\u03a9; $R_{4}^{{}}$ = 10\u03a9. \u0110\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf U th\u00ec \u0111o \u0111\u01b0\u1ee3c hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ l\u00e0 2V. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf U.","select":["A. 6V","B. 12V","C. 18V","D. 24V"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch: $I=I_{1}^{{}}=\\dfrac{U_{1}^{{}}}{R_{1}^{{}}}=\\dfrac{2}{2}=1A$<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng: $R=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}+R_{4}^{{}}=2+4+8+10=24\\Omega $<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf hai \u0111\u1ea7u m\u1ea1ch: $U=\\text{I}\\times \\text{R}=24V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2926},{"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/bai6/lv2/img\/h17.jpg'\/><\/center>Cho \u0111o\u1ea1n m\u1ea1ch \u0111i\u1ec7n theo s\u01a1 \u0111\u1ed3 nh\u01b0 h\u00ecnh v\u1ebd, trong \u0111\u00f3 \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 5\u2126; $R_{2}^{{}}$ = 15\u2126; v\u00f4n k\u1ebf ch\u1ec9 3V. T\u00ednh hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch AB.","select":["A. 4V","B. 12V","C. 18V","D. 24V"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n: $I=\\dfrac{U_{2}^{{}}}{R_{2}^{{}}}=\\dfrac{3}{15}=0,2A$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch: $U=\\text{I}\\times \\text{R}=0,2\\times (5+15)=4V$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2927},{"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":"C\u00f3 b\u1ed1n \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ = 15\u03a9; $R_{2}^{{}}$ = 25\u03a9; $R_{3}^{{}}$ = 20\u03a9; $R_{4}^{{}}$ = 30\u03a9. M\u1eafc b\u1ed1n \u0111i\u1ec7n tr\u1edf n\u00e0y n\u1ed1i ti\u1ebfp v\u1edbi nhau r\u1ed3i \u0111\u1eb7t v\u00e0o hai \u0111\u1ea7u \u0111o\u1ea1n m\u1ea1ch hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 90V. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch l\u00e0:","select":["A. I = 2A","B. I = 1,5A","C. I = 1A","D. I = 4,5A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng: $R_{td}^{{}}=R_{1}^{{}}+R_{2}^{{}}+R_{3}^{{}}+R_{4}^{{}}=15+25+20+30=90\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n trong m\u1ea1ch: $I=\\dfrac{U}{R_{td}^{{}}}=\\dfrac{90}{90}=1A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2928},{"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":"C\u00f4ng th\u1ee9c n\u00e0o sau \u0111\u00e2y l\u00e0 c\u00f4ng th\u1ee9c t\u00ednh \u0111i\u1ec7n tr\u1edf m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp.","select":["A. $R=\\dfrac{1}{R_{1}^{{}}}+\\dfrac{1}{R_{2}^{{}}}$","B. $R=R_{1}^{{}}+R_{2}^{{}}$","C. $\\dfrac{1}{R}=\\dfrac{1}{R_{1}^{{}}}+\\dfrac{1}{R_{2}^{{}}}$","D. $R=\\dfrac{R_{1}^{{}}R_{2}^{{}}}{R_{1}^{{}}+R_{2}^{{}}}$"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c t\u00ednh \u0111i\u1ec7n tr\u1edf m\u1ea1ch m\u1eafc n\u1ed1i ti\u1ebfp l\u00e0 $R=R_{1}^{{}}+R_{2}^{{}}$.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2929},{"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\u1eafc n\u1ed1i ti\u1ebfp $R_{1}^{{}}$ = 10\u03a9 v\u00e0 $R_{2}^{{}}$ = 30\u03a9 v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf kh\u00f4ng \u0111\u1ed5i U = 12V. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111i\u1ec7n tr\u1edf $R_{1}^{{}}$ l\u00e0","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: $R=R_{1}^{{}}+R_{2}^{{}}=10+30=40\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n: $I_{1}^{{}}=I=\\dfrac{U}{R}=\\dfrac{12}{40}=0,3A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2930}],"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)

Điểm của bạn.

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