{"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":"Tr\u00ean b\u00e0n l\u00e0 c\u00f3 ghi 220V-1100W. Khi b\u00e0n l\u00e0 n\u00e0y ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng th\u00ec n\u00f3 c\u00f3 \u0111i\u1ec7n tr\u1edf l\u00e0 bao nhi\u00eau?","select":["A. 0,2\u2126","B. 5\u2126","C. 44\u2126","D. 5500\u2126"],"hint":"","explain":"<span class='basic_left'>Khi b\u00e0n l\u00e0 n\u00e0y ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng th\u00ec n\u00f3 c\u00f3 \u0111i\u1ec7n tr\u1edf l\u00e0 :<br\/>$P=\\dfrac{{{U}^{2}}}{R}\\Rightarrow R=\\dfrac{{{U}^{2}}}{P}=\\dfrac{{{220}^{2}}}{1100}=44\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2871},{"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":"Tr\u00ean nhi\u1ec1u d\u1ee5ng c\u1ee5 \u0111i\u1ec7n trong gia \u0111\u00ecnh th\u01b0\u1eddng c\u00f3 ghi 220V v\u00e0 s\u1ed1 o\u00e1t(W), s\u1ed1 o\u00e1t n\u00e0y c\u00f3 \u00fd ngh\u0129a l\u00e0:","select":["A. C\u00f4ng su\u1ea5t ti\u00eau th\u1ee5 \u0111i\u1ec7n c\u1ee7a d\u1ee5ng c\u1ee5 khi n\u00f3 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng v\u1edbi nh\u1eefng hi\u1ec7u \u0111i\u1ec7n th\u1ebf nh\u1ecf h\u01a1n 220V.","B. C\u00f4ng su\u1ea5t ti\u00eau th\u1ee5 \u0111i\u1ec7n c\u1ee7a d\u1ee5ng c\u1ee5 khi n\u00f3 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng v\u1edbi \u0111\u00fang hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V.","C. C\u00f4ng m\u00e0 d\u00f2ng \u0111i\u1ec7n th\u1ef1c hi\u1ec7n trong m\u1ed9t ph\u00fat khi d\u1ee5ng c\u1ee5 n\u00e0y \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng v\u1edbi \u0111\u00fang hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V.","D. \u0110i\u1ec7n n\u0103ng m\u00e0 d\u1ee5ng c\u1ee5 ti\u00eau th\u1ee5 trong m\u1ed9t gi\u1edd khi n\u00f3 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng v\u1edbi \u0111\u00fang hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V."],"hint":"","explain":"<span class='basic_left'>Tr\u00ean nhi\u1ec1u d\u1ee5ng c\u1ee5 \u0111i\u1ec7n trong gia \u0111\u00ecnh th\u01b0\u1eddng c\u00f3 ghi 220V v\u00e0 s\u1ed1 o\u00e1t(W), s\u1ed1 o\u00e1t n\u00e0y c\u00f3 \u00fd ngh\u0129a l\u00e0 c\u00f4ng su\u1ea5t ti\u00eau th\u1ee5 \u0111i\u1ec7n c\u1ee7a d\u1ee5ng c\u1ee5 khi n\u00f3 \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng v\u1edbi \u0111\u00fang hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V. <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2872},{"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 hai \u0111i\u1ec7n tr\u1edf ${{R}_{1}}$ v\u00e0 ${{R}_{2}}=2{{R}_{1}}$ \u0111\u01b0\u1ee3c m\u1eafc song song v\u00e0o m\u1ed9t hi\u1ec7u \u0111i\u1ec7n th\u1ebf kh\u00f4ng \u0111\u1ed5i. C\u00f4ng su\u1ea5t \u0111i\u1ec7n ${{P}_{1}}$, ${{P}_{2}}$ t\u01b0\u01a1ng \u1ee9ng tr\u00ean hai \u0111i\u1ec7n tr\u1edf n\u00e0y c\u00f3 m\u1ed1i quan h\u1ec7 n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?","select":["A. ${{P}_{1}}={{P}_{2}}$","B. ${{P}_{2}}=2{{P}_{1}}$","C. ${{P}_{1}}=2{{P}_{2}}$","D. ${{P}_{1}}=4{{P}_{2}}$"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng su\u1ea5t \u0111i\u1ec7n c\u1ee7a m\u1ea1ch \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c:<br\/> $P={{I}^{2}}\\times R=\\dfrac{{{U}^{2}}}{R}$<br\/>m\u00e0 ${{R}_{2}}=2{{R}_{1}}$ , n\u00ean ${{P}_{1}}=2{{P}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2873},{"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":"Tr\u00ean b\u00f3ng \u0111\u00e8n $\u0110_1$ c\u00f3 ghi 220V-100W, Tr\u00ean b\u00f3ng \u0111\u00e8n $\u0110_2$ c\u00f3 ghi 220V-25W. Khi \u0111\u00e8n s\u00e1ng b\u00ecnh th\u01b0\u1eddng, \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u1ee9ng ${{R}_{1}}$ v\u00e0 ${{R}_{2}}$ c\u1ee7a d\u00e2y t\u00f3c c\u00e1c b\u00f3ng \u0111\u00e8n n\u00e0y c\u00f3 m\u1ed1i quan h\u1ec7 v\u1edbi nhau nh\u01b0 th\u1ebf n\u00e0o?","select":["A. ${{R}_{1}}=4{{R}_{2}}$","B. $4{{R}_{1}}={{R}_{2}}$","C. ${{R}_{1}}=16{{R}_{2}}$","D. $16{{R}_{1}}={{R}_{2}}$"],"hint":"","explain":"<span class='basic_left'>Tr\u00ean b\u00f3ng \u0111\u00e8n $\u0110_1$ c\u00f3 ghi 220V-100W, tr\u00ean b\u00f3ng \u0111\u00e8n $\u00d0_2$ c\u00f3 ghi 220V-25W. <br\/>Ta th\u1ea5y ${{P}_{1}}=4{{P}_{2}}$ n\u00ean khi \u0111\u00e8n s\u00e1ng b\u00ecnh th\u01b0\u1eddng, \u0111i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u1ee9ng ${{R}_{1}}$ v\u00e0 ${{R}_{2}}$ c\u1ee7a d\u00e2y t\u00f3c c\u00e1c b\u00f3ng \u0111\u00e8n n\u00e0y c\u00f3 m\u1ed1i quan h\u1ec7 v\u1edbi nhau: $4{{R}_{1}}={{R}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2874},{"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 su\u1ea5t \u0111i\u1ec7n c\u1ee7a m\u1ed9t \u0111o\u1ea1n m\u1ea1ch c\u00f3 \u00fd ngh\u0129a g\u00ec?","select":["A. L\u00e0 n\u0103ng l\u01b0\u1ee3ng c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111o\u1ea1n m\u1ea1ch \u0111\u00f3.","B. L\u00e0 \u0111i\u1ec7n n\u0103ng m\u00e0 \u0111o\u1ea1n m\u1ea1ch \u0111\u00f3 ti\u00eau th\u1ee5 trong m\u1ed9t \u0111\u01a1n v\u1ecb th\u1eddi gian.","C. L\u00e0 m\u1ee9c \u0111\u1ed9 m\u1ea1nh y\u1ebfu c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111o\u1ea1n m\u1ea1ch \u0111\u00f3.","D. L\u00e0 c\u00e1c lo\u1ea1i t\u00e1c d\u1ee5ng m\u00e0 d\u00f2ng \u0111i\u1ec7n g\u00e2y ra \u1edf \u0111o\u1ea1n m\u1ea1ch."],"hint":"","explain":"<span class='basic_left'>C\u00f4ng su\u1ea5t \u0111i\u1ec7n c\u1ee7a m\u1ed9t \u0111o\u1ea1n m\u1ea1ch c\u00f3 \u00fd ngh\u0129a: L\u00e0 \u0111i\u1ec7n n\u0103ng m\u00e0 \u0111o\u1ea1n m\u1ea1ch \u0111\u00f3 ti\u00eau th\u1ee5 trong m\u1ed9t \u0111\u01a1n v\u1ecb th\u1eddi gian.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":2875},{"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":"Tr\u00ean m\u1ed9t b\u00f3ng \u0111\u00e8n c\u00f3 ghi 12V-6W. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 \u0111\u1ecbnh m\u1ee9c c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua \u0111\u00e8n.","select":["A. 0,8A","B. 0,75A","C. 0,25A","D. 0,5A"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $P=U\\times I\\Rightarrow I=\\dfrac{P}{U}=\\dfrac{6}{12}=0,5A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2876},{"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":"Tr\u00ean m\u1ed9t b\u00e0n l\u00e0 \u0111i\u1ec7n c\u00f3 ghi 12V-6W. T\u00ednh \u0111i\u1ec7n tr\u1edf c\u1ee7a b\u00e0n l\u00e0?","select":["A. 20\u03a9","B. 24\u03a9","C. 36\u03a9","D. 12\u03a9"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $P=U\\times I\\Rightarrow I=\\dfrac{P}{U}=\\dfrac{6}{12}=0,5A$<br\/>\u0110i\u1ec7n tr\u1edf c\u1ee7a b\u00e0n l\u00e0: $R=\\dfrac{{{U}^{2}}}{P}=\\dfrac{{{12}^{2}}}{6}=24\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2877},{"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 d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng ph\u1ea3i l\u00e0 c\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t ti\u00eau th\u1ee5 \u0111i\u1ec7n n\u0103ng P c\u1ee7a \u0111o\u1ea1n m\u1ea1ch \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U, d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 I v\u00e0 \u0111i\u1ec7n tr\u1edf c\u1ee7a n\u00f3 l\u00e0 R? ","select":["A. $P=U\\times I$","B. $P=\\dfrac{U}{I}$","C. $P=\\dfrac{{{U}^{2}}}{R}$","D. $P={{I}^{2}}\\times R$"],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2878},{"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":"Tr\u00ean m\u1ed9t b\u00f3ng \u0111\u00e8n \u0111i\u1ec7n c\u00f3 ghi 220V-528W. T\u00ednh c\u01b0\u1eddng \u0111\u1ed9 \u0111\u1ecbnh m\u1ee9c c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua b\u00f3ng \u0111\u00e8n:","select":["A. 4,4A","B. 1,5A","C. 2,4A","D. 1,2A"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 \u0111\u1ecbnh m\u1ee9c c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua b\u00f3ng \u0111\u00e8nl\u00e0:<br\/>Ta c\u00f3: $P=U\\times I\\Rightarrow I=\\dfrac{P}{U}=\\dfrac{528}{220}=2,4A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2879},{"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":"Tr\u00ean m\u1ed9t n\u1ed3i c\u01a1m \u0111i\u1ec7n c\u00f3 ghi 220V-528W. T\u00ednh \u0111i\u1ec7n tr\u1edf d\u00e2y nung c\u1ee7a n\u1ed3i khi n\u1ed3i \u0111ang ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng.","select":["A. 91,7\u03a9","B. 66,7\u03a9","C.75\u03a9","D.72\u03a9"],"hint":"","explain":"<span class='basic_left'>C\u01b0\u1eddng \u0111\u1ed9 \u0111\u1ecbnh m\u1ee9c c\u1ee7a d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua d\u00e2y nung c\u1ee7a n\u1ed3i l\u00e0:<br\/>Ta c\u00f3: $P=U\\times I\\Rightarrow I=\\dfrac{P}{U}=\\dfrac{528}{220}=2,4A$<br\/>\u0110i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y nung khi n\u1ed3i \u0111ang c\u00f2n ho\u1ea1t \u0111\u1ed9ng b\u00ecnh th\u01b0\u1eddng l\u00e0:<br\/> $R=\\dfrac{U}{I}=\\dfrac{220}{2,4}=91,7\\Omega $<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2880},{"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":"Tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_1$ c\u00f3 ghi 220V-100W, Tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_2$ c\u00f3 ghi 220V-75W. M\u1eafc song song hai b\u00f3ng \u0111\u00e8n n\u00e0y v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V. T\u00ednh c\u00f4ng su\u1ea5t c\u1ee7a \u0111o\u1ea1n m\u1ea1ch song song n\u00e0y.","select":["A. 225W","B. 120W","C. 150W","D. 175W"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y t\u00f3c b\u00f3ng \u0111\u00e8n $\u0110_1$ v\u00e0 $\u0110_2$ :<br\/>${{R}_{1}}=\\dfrac{{{U}_{1}}^{2}}{{{P}_{1}}}=\\dfrac{{{220}^{2}}}{100}=484\\Omega $<br\/>${{R}_{2}}=\\dfrac{{{U}_{2}}^{2}}{{{P}_{2}}}=\\dfrac{{{220}^{2}}}{75}=645,3\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf to\u00e0n m\u1ea1ch song song: $R=\\dfrac{{{R}_{1}}\\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\\dfrac{484\\times 645,3}{484+645,3}=276,3\\Omega $<br\/>C\u00f4ng su\u1ea5t c\u1ee7a \u0111o\u1ea1n m\u1ea1ch : $P=\\dfrac{{{U}^{2}}}{R}=\\dfrac{{{220}^{2}}}{276,6}=175W$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":2881},{"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":"Tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_1$ c\u00f3 ghi 220V-100W, Tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_2$ c\u00f3 ghi 220V-75W. M\u1eafc hai \u0111\u00e8n tr\u00ean \u0111\u00e2y n\u1ed1i ti\u1ebfp v\u1edbi nhau r\u1ed3i m\u1eafc \u0111o\u1ea1n m\u1ea1ch n\u00e0y v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V. T\u00ednh c\u00f4ng su\u1ea5t \u0111i\u1ec7n c\u1ee7a \u0111o\u1ea1n m\u1ea1ch n\u1ed1i ti\u1ebfp n\u00e0y, cho r\u1eb1ng \u0111i\u1ec7n tr\u1edf c\u1ee7a m\u1ed7i \u0111\u00e8n khi \u0111\u00f3 b\u1eb1ng 50% \u0111i\u1ec7n tr\u1edf c\u1ee7a \u0111\u00e8n \u0111\u00f3 khi s\u00e1ng b\u00ecnh th\u01b0\u1eddng.","select":["A. 85,8W","B. 33,3W","C. 66,7W","D. 85W"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y t\u00f3c b\u00f3ng \u0111\u00e8n $\u0110_1$ v\u00e0 $\u0110_2$ : <br\/>${{R}_{1}}=\\dfrac{{{U}_{1}}^{2}}{{{P}_{1}}}=\\dfrac{{{220}^{2}}}{100}=484\\Omega $<br\/>${{R}_{2}}=\\dfrac{{{U}_{2}}^{2}}{{{P}_{2}}}=\\dfrac{{{220}^{2}}}{75}=645,3\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a to\u00e0n m\u1ea1ch n\u1ed1i ti\u00eap: $R={{R}_{1}}+{{R}_{2}}=484+645,3=1129,3\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch: $I=\\dfrac{U}{R}=\\dfrac{220}{1129,3}\\approx 0,195A={{I}_{1}}={{I}_{2}}$<br\/>Hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u \u0111\u00e8n $\u0110_1$ v\u00e0 $\u0110_2$ :<br\/>${{U}_{1}}=I\\times {{R}_{1}}=0,195\\times 484=94,38V$<br\/>${{U}_{2}}=I\\times {{R}_{2}}=0,195\\times 645,3=125,83V$<br\/>C\u00f4ng su\u1ea5t c\u1ee7a \u0111o\u1ea1n m\u1ea1ch: ${{P}_{1}}=\\dfrac{{{U}_{1}}^{2}}{{{R}_{1}}}=\\dfrac{94,{{38}^{2}}}{\\dfrac{484}{2}}=36,8W$<br\/>${{P}_{2}}=\\dfrac{{{U}_{2}}^{2}}{{{R}_{2}}}=\\dfrac{125,{{83}^{2}}}{645,3}=49W$<br\/>$\\Rightarrow P={{P}_{1}}+{{P}_{2}}=36,8+49=85,8W$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2882},{"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":"M\u1ed9t b\u1ebfp \u0111i\u1ec7n c\u00f3 \u0111i\u1ec7n tr\u1edf R \u0111\u01b0\u1ee3c m\u1eafc v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf U th\u00ec d\u00f2ng \u0111i\u1ec7n ch\u1ea1y qua n\u00f3 c\u00f3 c\u01b0\u1eddng \u0111\u1ed9 I. Khi \u0111\u00f3 c\u00f4ng su\u1ea5t c\u1ee7a b\u1ebfp l\u00e0 P. C\u00f4ng th\u1ee9c P n\u00e0o d\u01b0\u1edbi \u0111\u00e2y kh\u00f4ng \u0111\u00fang?","select":["A. $P={{U}^{2}}\\times R$","B. $P=\\dfrac{{{U}^{2}}}{R}$","C. $P={{I}^{2}}\\times R$","D. $P=U\\times I$"],"hint":"","explain":"<span class='basic_left'><br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2883},{"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":"C\u00f3 tr\u01b0\u1eddng h\u1ee3p, khi b\u00f3ng \u0111\u00e8n b\u1ecb \u0111\u1ee9t d\u00e2y t\u00f3c, ta c\u00f3 th\u1ec3 l\u1eafc cho hai \u0111\u1ea7u d\u00e2y t\u00f3c \u1edf ch\u1ed7 b\u1ecb \u0111\u1ee9t d\u00ednh l\u1ea1i v\u1edbi nhau v\u00e0 c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng b\u00f3ng \u0111\u00e8n n\u00e0y th\u00eam m\u1ed9t th\u1eddi gian n\u1eefa. H\u1ecfi khi \u0111\u00f3 c\u00f4ng su\u1ea5t v\u00e0 \u0111\u1ed9 s\u00e1ng c\u1ee7a b\u00f3ng \u0111\u00e8n l\u1edbn h\u01a1n hay nh\u1ecf h\u01a1n so v\u1edbi tr\u01b0\u1edbc khi d\u00e2y t\u00f3c b\u1ecb \u0111\u1ee9t?","select":["A. L\u1edbn h\u01a1n","B. Nh\u1ecf h\u01a1n","C. B\u1eb1ng nhau","D. Kh\u00f4ng so s\u00e1nh \u0111\u01b0\u1ee3c"],"hint":"","explain":"<span class='basic_left'>Khi b\u1ecb \u0111\u1ee9t v\u00e0 \u0111\u01b0\u1ee3c n\u1ed1i d\u00ednh l\u1ea1i th\u00ec d\u00e2y t\u00f3c c\u1ee7a b\u00f3ng \u0111\u00e8n ng\u1eafn h\u01a1n tr\u01b0\u1edbc n\u00ean \u0111i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y t\u00f3c nh\u1ecf h\u01a1n tr\u01b0\u1edbc. Trong khi \u0111\u00f3, hi\u1ec7u \u0111i\u1ec7n th\u1ebf gi\u1eefa hai \u0111\u1ea7u d\u00e2y t\u00f3c v\u1eabn nh\u01b0 tr\u01b0\u1edbc n\u00ean c\u00f4ng su\u1ea5t $P=\\dfrac{{{U}^{2}}}{R}$s\u1ebd l\u1edbn h\u01a1n. Do v\u1eady \u0111\u00e8n s\u1ebd s\u00e1ng h\u01a1n so v\u1edbi tr\u01b0\u1edbc.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2884},{"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":"Tr\u00ean hai b\u00f3ng \u0111\u00e8n c\u00f3 ghi 220V \u2013 60W v\u00e0 220V \u2013 75W. Bi\u1ebft r\u1eb1ng d\u00e2y t\u00f3c c\u1ee7a hai b\u00f3ng \u0111\u00e8n n\u00e0y \u0111\u1ec1u b\u1eb1ng vonfam v\u00e0 c\u00f3 ti\u1ebft di\u1ec7n b\u1eb1ng nhau. D\u00e2y t\u00f3c c\u1ee7a \u0111\u00e8n n\u00e0o c\u00f3 \u0111\u1ed9 d\u00e0i l\u1edbn h\u01a1n v\u00e0 l\u1edbn h\u01a1n bao nhi\u00eau l\u1ea7n?","select":["A. 0,5","B. 1","C. 1,25","D. 2,5"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf c\u1ee7a b\u00f3ng \u0111\u00e8n 60W l\u00e0: ${{R}_{1}}=\\dfrac{{{220}^{2}}}{60}$ <br\/>\u0110i\u1ec7n tr\u1edf c\u1ee7a b\u00f3ng \u0111\u00e8n 75W l\u00e0: ${{R}_{2}}=\\dfrac{{{220}^{2}}}{75}$ <br\/>M\u00e0 \u0111i\u1ec7 tr\u1edf t\u1ef7 l\u1ec7 thu\u1eadn v\u1edbi chi\u1ec1u d\u00e0i, n\u00ean: $\\dfrac{{{l}_{1}}}{{{l}_{2}}}=\\dfrac{{{R}_{1}}}{{{R}_{2}}}=\\dfrac{75}{60}=1,25$ <br\/>V\u1eady d\u00e2y t\u00f3c c\u1ee7a b\u00f3ng \u0111\u00e8n 60W s\u1ebd d\u00e0i h\u01a1n v\u00e0 d\u00e0i h\u01a1n 1,25l\u1ea7n.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2885},{"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":"M\u1eafc m\u1ed9t b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c c\u00f3 ghi 220V \u2013 60W v\u00e0o \u1ed5 l\u1ea5y \u0111i\u1ec7n c\u00f3 hi\u1ec7u \u0111i\u1ec7n th\u1ebf 110V. Cho r\u1eb1ng \u0111i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y t\u00f3c b\u00f3ng \u0111\u00e8n kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o nhi\u1ec7t \u0111\u1ed9, t\u00ednh c\u00f4ng su\u1ea5t c\u1ee7a b\u00f3ng \u0111\u00e8n khi \u0111\u00f3?","select":["A. 15W","B. 33,3W","C. 66,7W","D. 85W"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t: $P=\\dfrac{{{U}^{2}}}{{{R}_{d}}}\\Rightarrow {{R}_{d}}=\\dfrac{{{U}^{2}}}{P}=\\dfrac{{{220}^{2}}}{60}=806,67\\Omega $<br\/>V\u00ec \u0111i\u1ec7n tr\u1edf R c\u1ee7a \u0111\u00e8n kh\u00f4ng \u0111\u1ed5i, n\u00ean khi m\u1eafc \u0111\u00e8n v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 110V th\u00ec \u0111\u00e8n ch\u1ea1y v\u1edbi c\u00f4ng su\u1ea5t: $P=\\dfrac{U{{'}^{2}}}{{{R}_{d}}}=\\dfrac{{{110}^{2}}}{806,67}=15W$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2886},{"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":"\u1ede c\u00f4ng tr\u01b0\u1eddng x\u00e2y d\u1ef1ng c\u00f3 s\u1eed d\u1ee5ng m\u1ed9t m\u00e1y n\u00e2ng \u0111\u1ec3 n\u00e2ng kh\u1ed1i v\u1eadt li\u1ec7u c\u00f3 tr\u1ecdng l\u01b0\u1ee3ng 2000N l\u00ean t\u1edbi \u0111\u1ed9 cao 15m trong th\u1eddi gian 40 gi\u00e2y. Ph\u1ea3i d\u00f9ng \u0111\u1ed9ng c\u01a1 \u0111i\u1ec7n c\u00f3 c\u00f4ng su\u1ea5t n\u00e0o d\u01b0\u1edbi \u0111\u00e2y l\u00e0 th\u00edch h\u1ee3p cho m\u00e1y n\u00e2ng n\u00e0y.","select":["A. 120kW","B. 0,75kW","C. 75W","D. 7,5kW"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng su\u1ea5t c\u1ee7a m\u00e1y n\u00e2ng:$P=\\dfrac{A}{t}=\\dfrac{2000\\times 15}{40}=750W=0,75kW$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":2887},{"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":"Tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_1$ c\u00f3 ghi 220V \u2013 100W, tr\u00ean b\u00f3ng \u0111\u00e8n d\u00e2y t\u00f3c $\u0110_2$ c\u00f3 ghi 220V \u2013 75W. M\u1eafc hai \u0111\u00e8n tr\u00ean d\u00e2y n\u1ed1i ti\u1ebfp v\u1edbi nhau r\u1ed3i m\u1eafc \u0111o\u1ea1n m\u1ea1ch n\u00e0y v\u00e0o hi\u1ec7u \u0111i\u1ec7n th\u1ebf 220V. T\u00ednh d\u00f2ng \u0111i\u1ec7n (I) \u0111i qua \u0111o\u1ea1n m\u1ea1ch.","select":["A. 0,195A","B. 0,39A","C. 0,78A","D. 1,95A"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec7n tr\u1edf c\u1ee7a d\u00e2y t\u00f3c b\u00f3ng \u0111\u00e8n $\u0110_1$ v\u00e0 $\u0110_2$ :<br\/>${{R}_{1}}=\\dfrac{{{U}_{1}}^{2}}{{{P}_{1}}}=\\dfrac{{{220}^{2}}}{100}=484\\Omega $<br\/>${{R}_{2}}=\\dfrac{{{U}_{2}}^{2}}{{{P}_{2}}}=\\dfrac{{{220}^{2}}}{75}=645,3\\Omega $<br\/>\u0110i\u1ec7n tr\u1edf t\u01b0\u01a1ng \u0111\u01b0\u01a1ng c\u1ee7a to\u00e0n m\u1ea1ch n\u1ed1i ti\u1ebfp: $R={{R}_{1}}+{{R}_{2}}=484+645,3=1129,3\\Omega $<br\/>C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua m\u1ea1ch: $I=\\dfrac{U}{R}=\\dfrac{220}{1129,3}\\approx 0,195A={{I}_{1}}={{I}_{2}}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2888},{"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":"Tr\u00ean m\u1ed9t b\u00f3ng \u0111\u00e8n c\u00f3 ghi 6V \u2013 3W. C\u01b0\u1eddng \u0111\u1ed9 d\u00f2ng \u0111i\u1ec7n qua b\u00f3ng khi n\u00f3 sang b\u00ecnh th\u01b0\u1eddng l\u00e0 bao nhi\u00eau?","select":["A. 0,5A","B. 2A","C. 18A","D. 12A"],"hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $P=U\\times I\\Rightarrow I=\\dfrac{P}{U}=\\dfrac{3}{6}=0,5A$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":2889},{"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":" \u0110i\u1ec7n tr\u1edf c\u1ee7a b\u1ebfp \u0111i\u1ec7n l\u00e0m b\u1eb1ng nik\u00ealin R = 48,5\u03a9. B\u1ebfp \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u1edf hi\u1ec7u \u0111i\u1ec7n th\u1ebf U = 220V. C\u00f4ng su\u1ea5t ti\u00eau th\u1ee5 c\u1ee7a b\u1ebfp \u0111i\u1ec7n g\u1ea7n \u0111\u00fang nh\u1ea5t l\u00e0:","select":["A. 99,79W","B. 9,979W","C. 997,9W","D. 0,9979W"],"hint":"","explain":"<span class='basic_left'>C\u00f4ng th\u1ee9c t\u00ednh c\u00f4ng su\u1ea5t: $P=\\dfrac{{{U}^{2}}}{R}=\\dfrac{{{220}^{2}}}{48,5}=997,9W$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":2890}],"lesson":{"save":0,"level":2}}