{"event_info":{"picture":"https://img.loigiaihay.com/picture/event/2018/1106/45p-c3-0.jpg","title":"Đề kiểm tra 45 phút (1 tiết) –  Chương 3 – Vật lí 12","display_type":1,"cat_id":63,"cat_title":"SGK Vật lí lớp 12","alias":"de-kiem-tra-45-phut-1-tiet-chuong-3-vat-li-12-co-ban","introtext":"","content_box_above":"","type":"event","page_builder":"","item_id":10005,"item_type":2},"list_articles":[{"article_id":46709,"cat_id":63,"publish_time":1541475060,"type_ex":2,"is_hot":0,"is_link":0,"thumbnail":"https://img.loigiaihay.com/picture/article/2018/1106/resize_de-so-1-45p-c3-0.jpg","title":"Đề kiểm tra 45 phút (1 tiết) – Đề số 1 -  Chương 3 – Vật lí 12","label":"","alias":"de-kiem-tra-45-phut-1-tiet-de-so-1-chuong-3-vat-li-12-co-ban","introtext":"Đáp án và lời giải chi tiết Đề kiểm tra 45 phút (1 tiết) – Đề số 1 -  Chương 3 – Vật lí 12","display_type":1,"article_type":1,"content":"<style>\r\n.method-close-icon{position: relative;float: right;top: 6px;transform: rotate(270deg);-ms-transform:rotate(270deg);-moz-transform:rotate(270deg);-webkit-transform:rotate(270deg);-o-transform:rotate(270deg);}\r\n.method-open-icon{position: relative;float: right;top: 6px;}\r\n</style>\r\n<p><strong style=\"font-size: 16px;\">Lời giải chi tiết</strong></p><p style=\"text-align: justify;\"><strong>Đáp án</strong></p>\r\n<table border=\"1\" cellspacing=\"0\" cellpadding=\"0\">\r\n<tbody>\r\n<tr>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>1. A</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>2. D</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>3. B</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>4. D</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>5. B</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>6. C</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>7. A</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>8. A</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>9. C</strong></p>\r\n</td>\r\n</tr>\r\n<tr>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>10. B</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>11. B</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>12. A</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>13. D</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>14. A</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>15. C</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>16. B</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>17. D</strong></p>\r\n</td>\r\n<td valign=\"top\" width=\"71\">\r\n<p><strong>18. A</strong></p>\r\n</td>\r\n</tr>\r\n</tbody>\r\n</table>\r\n<p style=\"text-align: justify;\"><strong>Giải chi tiết</strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 1: A </strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 2: D</strong></p>\r\n<p><strong>Câu 3:</strong></p>\r\n<p>Độ lệch pha giữa u và i là:</p>\r\n<p>\\(\\varphi  = {\\varphi _u} - {\\varphi _i} = \\frac{\\pi }{4} - \\left( { - \\frac{\\pi }{4}} \\right) = \\frac{\\pi }{2}\\)</p>\r\n<p>=&gt; u sớm pha hơn i góc \\(\\frac{\\pi }{2}\\) =&gt; Mạch chứa cuộn cảm thuần.</p>\r\n<p><strong>Chọn B</strong></p>\r\n<p><strong>Câu 4:</strong></p>\r\n<p>\\({P_{\\max }}\\)khi xảy ra cộng hưởng \\( \\Leftrightarrow {Z_L} = {Z_C} \\Leftrightarrow L\\omega  = \\frac{1}{{C\\omega }} \\Leftrightarrow \\omega  = \\sqrt {\\frac{1}{{LC}}} \\)</p>\r\n<p><strong>Chọn D </strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 5: B</strong></p>\r\n<p><strong>Câu 6:</strong></p>\r\n<p>R thay đổi, P max \\( \\Leftrightarrow R = \\left| {{Z_L} - {Z_C}} \\right| = \\left| {80 - 30} \\right| = 50\\)</p>\r\n<p><strong>Chọn C</strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 7: A</strong></p>\r\n<p style=\"text-align: justify;\">Ta có:</p>\r\n<p style=\"text-align: justify;\">\\({U_C} = {U_{AM}};{U_L} = {U_{MN}};\\)\\(\\,{U_R} = {U_{NB}};{U_{RL}} = {U_{MB}}\\)</p>\r\n<p style=\"text-align: justify;\">Do đó: \\(\\begin{array}{l}U_{RL}^2 = U_R^2 + U_L^2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1)\\\\U_{AB}^2 = U_R^2 + U_L^2 + U_C^2 - 2{U_L}.{U_C}\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Từ (1) và (2) ta suy ra: \\(U_C^2 = 2{U_L}.{U_C} \\Rightarrow {U_L} = \\dfrac{{{U_C}}}{2} = 20V\\)</p>\r\n<p style=\"text-align: justify;\">Thay U<sub>L</sub> vào (1) ta có: U<sub>R</sub>=20V</p>\r\n<p style=\"text-align: justify;\">Vậy \\(cos\\varphi  = \\dfrac{{{U_R}}}{{{U_{AB}}}} = \\dfrac{{20}}{{20\\sqrt 2 }} = \\dfrac{{\\sqrt 2 }}{2}\\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 8: A</strong></p>\r\n<p style=\"text-align: justify;\">Từ công thức: \\(f = \\dfrac{{np}}{{60}} \\Rightarrow n = \\dfrac{{60f}}{p} = 300\\)  vòng/phút</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 9: C</strong></p>\r\n<p style=\"text-align: justify;\">Vì \\(\\begin{array}{l}\\tan \\varphi  = \\dfrac{{{U_L} - {U_C}}}{{{U_R}}} = \\tan \\dfrac{\\pi }{4} = 1\\\\ \\Rightarrow {U_L} - {U_C} = {U_R};\\\\P = R{I^2},U_{AB}^2 = U_R^2 + {({U_L} - {U_C})^2};\\\\{Z_C} &gt; {Z_L}\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Dùng phép thử ta có: \\(R = 50\\Omega ;{Z_C} = 100\\Omega \\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 10: B</strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 11: B</strong></p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l}{Z_C} = \\dfrac{1}{{\\omega C}} = \\dfrac{1}{{100\\pi \\dfrac{{{{10}^{ - 4}}}}{{3\\pi }}}} = 300\\Omega \\\\Z = \\sqrt {{R^2} + Z_C^2}  = \\dfrac{U}{I} = \\dfrac{{150}}{{\\dfrac{{\\sqrt 5 }}{5}}} = 150\\sqrt 5 \\Omega \\\\ \\Rightarrow R = 150\\Omega \\end{array}\\)<strong> </strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 12: A</strong></p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l}{Z_{AN}} = \\sqrt {{R^2} + Z_C^2}  = \\dfrac{{{U_{AN}}}}{I} = \\dfrac{{200}}{2} = 100\\Omega \\\\ \\Rightarrow {R^2} + Z_C^2 = {100^2}\\,\\,\\,\\,\\,\\,(1)\\\\{Z_{MB}} = \\sqrt {{R^2} + Z_L^2}  = \\dfrac{{{U_{MB}}}}{I} = \\dfrac{{150}}{2} = 75\\Omega\\\\  \\Rightarrow {R^2} + Z_L^2 = {75^2}\\,\\,\\,\\,\\,\\,\\,\\,\\,(2)\\end{array}\\)<strong> </strong></p>\r\n<p style=\"text-align: justify;\">Vì u<sub>AM</sub> vuông pha với u<sub>NB</sub>, nên ta có:</p>\r\n<p style=\"text-align: justify;\">\\(\\tan {\\varphi _{\\left( {{{{U_{AN}}} \\over i}} \\right)}}.\\tan {\\varphi _{\\left( {{{{U_{MB}}} \\over i}} \\right)}} = 1\\)</p>\r\n<p style=\"text-align: justify;\">\\( \\Leftrightarrow \\dfrac{{{Z_C}}}{R}.\\dfrac{{{Z_L}}}{R} = 1 \\Rightarrow {Z_C}{Z_L} = {R^2}\\,\\,\\,\\,\\,\\,\\,\\,\\,(3)\\)</p>\r\n<p style=\"text-align: justify;\">Từ (1), (2), (3) ta suy ra: \\({Z_L} = 45\\Omega .\\)  Độ tự cảm có giá trị:</p>\r\n<p style=\"text-align: justify;\">\\(L = \\dfrac{{{Z_L}}}{\\omega } = \\dfrac{{0,45}}{\\pi }H\\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 13: D</strong></p>\r\n<p style=\"text-align: justify;\">Vì khi L=L<sub>1</sub> hoặc L=L<sub>2</sub> thì i<sub>1</sub>=i<sub>2</sub>=i nên ta có: Z<sub>1</sub>=Z<sub>2</sub></p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l} \\Leftrightarrow \\sqrt {{R^2} + {{({Z_{{L_1}}} - {Z_C})}^2}}  = \\sqrt {{R^2} + {{({Z_{{L_2}}} - {Z_C})}^2}} \\\\ \\Rightarrow ({Z_{{L_1}}} - {Z_C}) =  \\pm ({Z_{{L_1}}} - {Z_C}) \\\\ \\Rightarrow ({Z_{{L_1}}} - {Z_C}) =  - ({Z_{{L_1}}} - {Z_C})\\\\ \\Rightarrow {Z_C} = \\dfrac{{{Z_{{L_1}}} + {Z_{{L_2}}}}}{2} = \\dfrac{{300\\sqrt 3  + 100\\sqrt 3 }}{2} \\\\= 200\\sqrt 3 \\Omega \\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Mặc khác ta lại có:</p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l}\\tan ({\\varphi _1} + {\\varphi _2}) = \\dfrac{{\\tan {\\varphi _1} + \\tan {\\varphi _2}}}{{1 - \\tan {\\varphi _1}.\\tan {\\varphi _2}}} \\\\= \\tan \\dfrac{{2\\pi }}{3} =  - \\sqrt 3 \\\\ \\Leftrightarrow \\dfrac{{\\dfrac{{{Z_{{L_1}}} - {Z_C}}}{R} + \\dfrac{{{Z_C} - {Z_{{L_2}}}}}{R}}}{{1 - ({Z_{{L_1}}} - {Z_C})({Z_{{L_1}}} - {Z_C})}} \\\\= \\dfrac{{200\\sqrt 3 R}}{{{R^2} - 100\\sqrt 3 .100\\sqrt 3 }} =  - \\sqrt 3 \\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Từ đó ta suy ra: \\(R = 100\\Omega \\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 14: A</strong></p>\r\n<p style=\"text-align: justify;\">Vì u sớm pha hơn I nên đoạn mạch gồm hai phần tử là điện trở thuần và cuộn cảm thuần.Do đó ta có:</p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l}Z = \\sqrt {{R^2} + Z_L^2}  = \\dfrac{U}{I} = \\dfrac{{120\\sqrt 2 }}{{1,2}} = 100\\sqrt 2 \\\\\\tan \\varphi  = \\dfrac{{{Z_L}}}{R} = \\tan \\dfrac{\\pi }{4} = 1\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Từ đó ta có: \\(R = {Z_L} = 100\\Omega .\\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 15: C</strong></p>\r\n<p style=\"text-align: justify;\">Vì \\(\\begin{array}{l}\\tan \\varphi  = \\dfrac{{{U_C} - {U_L}}}{{{U_R}}} = \\tan \\dfrac{\\pi }{4} = 1 \\\\ \\Rightarrow {U_C} - {U_L} = {U_R}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1)\\\\u = \\sqrt {U_R^2 + {{({U_L} - {U_C})}^2}}  = \\sqrt {2{{({U_L} - {U_C})}^2}}\\\\  \\Rightarrow {U_L} - {U_C} = 50\\sqrt 2 \\,\\,\\,\\,\\,\\,\\,(2)\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Từ (1) và (2) suy ra: \\({U_C} = 100\\sqrt 2 V;{U_R} = 50\\sqrt 2 V\\)</p>\r\n<p style=\"text-align: justify;\">Từ công thức:</p>\r\n<p style=\"text-align: justify;\">\\(P = {U_R}I \\Rightarrow I = \\dfrac{P}{{{U_R} = \\dfrac{{100}}{{50\\sqrt 2 }}}} = \\sqrt 2 A\\)</p>\r\n<p style=\"text-align: justify;\">Vậy \\({Z_C} = \\dfrac{{{U_C}}}{I} = \\dfrac{{100\\sqrt 2 }}{{\\sqrt 2 }} = 100\\Omega\\)</p>\r\n<p style=\"text-align: justify;\">\\(  \\Rightarrow C = \\dfrac{1}{{\\omega {Z_C}}} = \\dfrac{{{{10}^{ - 4}}}}{\\pi }F\\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 16: B</strong></p>\r\n<p style=\"text-align: justify;\">Ta có: \\(\\begin{array}{l}P = R{I^2} = R\\dfrac{{{U^2}}}{{{R^2} + {{({Z_L} - {Z_C})}^2}}}\\\\ \\Rightarrow P{R^2} - {U^2}R + P{({Z_L} - {Z_C})^2} = 0\\\\ \\Leftrightarrow 3{R^2} - 25R + 12 = 0\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Giải phương trình ta có: \\(R = 60\\Omega \\,hay\\,R = 106,7\\Omega \\)</p>\r\n<p style=\"text-align: justify;\"><strong>Câu 17: D</strong></p>\r\n<p><strong>Câu 18:</strong></p>\r\n<p>Ta có:</p>\r\n<p>\\(\\frac{{{U_1}}}{{{U_2}}} = \\frac{{{N_1}}}{{{N_2}}} = \\frac{{{I_2}}}{{{I_1}}} \\Leftrightarrow \\frac{{220}}{{{U_2}}} = \\frac{{{I_2}}}{2} = \\frac{{2{N_2}}}{{{N_2}}}\\)</p>\r\n<p>\\( \\Rightarrow \\left\\{ \\begin{array}{l}{I_2} = 4{\\rm{A}}\\\\{U_2} = 110V\\end{array} \\right.\\)</p>\r\n<p><strong>Chọn A</strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 19:</strong></p>\r\n<p style=\"text-align: justify;\">\\(P = UIcos\\varphi  = \\dfrac{{400}}{{\\sqrt 2 }}.\\dfrac{1}{{\\sqrt 2 }}.cos\\dfrac{\\pi }{6} = 100\\sqrt 3 {\\rm{W}}.\\)<strong> </strong></p>\r\n<p style=\"text-align: justify;\"><strong>Câu 20:</strong></p>\r\n<p style=\"text-align: justify;\">\\(\\begin{array}{l}{Z_C} = \\dfrac{1}{{\\omega C}} = \\dfrac{1}{{100\\pi .31,8}} = 100\\Omega ;\\\\{Z_L} = \\omega L = \\dfrac{{100\\pi .0,7}}{\\pi } = 70\\Omega \\\\Z = \\sqrt {{R^2} + {{({Z_L} - {Z_C})}^2}}  = \\sqrt {{{100}^2} + {{(100 - 70)}^2}}  \\\\= 104\\Omega \\\\I = \\dfrac{U}{Z} = \\dfrac{{100}}{{104}} = 0,96A;\\\\ \\tan \\varphi  = \\dfrac{{{Z_L} - {Z_C}}}{R} = \\dfrac{{70 - 100}}{{100}} = 0,3\\\\ \\Rightarrow \\varphi  = 0,29\\,rad\\end{array}\\)</p>\r\n<p style=\"text-align: justify;\">Biểu thức của cường độ dòng điện qua mạch là:</p>\r\n<p style=\"text-align: justify;\">\\(i = 0,96\\sqrt 2 cos(100\\pi t + 0,29)\\,(A)\\)</p>\r\n<p style=\"text-align: right;\"><strong>Loigiaihay.com</strong></p>"}],"most_views":[{"article_id":46709,"cat_id":63,"publish_time":1541475060,"is_hot":0,"is_link":0,"thumbnail":"https://img.loigiaihay.com/picture/article/2018/1106/resize_de-so-1-45p-c3-0.jpg","title":"Đề kiểm tra 45 phút (1 tiết) – Đề số 1 -  Chương 3 – Vật lí 12","alias":"de-kiem-tra-45-phut-1-tiet-de-so-1-chuong-3-vat-li-12-co-ban","introtext":"Đáp án và lời giải chi tiết Đề kiểm tra 45 phút (1 tiết) – Đề số 1 -  Chương 3 – Vật lí 12","display_type":1}],"list_sub_event":[],"top_banner":"","cfg_info":{"last_cfg_time":"1521866679","ad.popup_frequency":"0","ad.adsense.detail_bottom.unit_id":"ca-app-pub-8529835372050931/9927553003","ad.adsense.popup.unit_id":"ca-app-pub-8529835372050931/2161779402","current.force.update.version":"181","support.email":"loigiaihay.com@gmail.com","support.fanpage":"https://www.facebook.com/Loigiaihay/","popup_rating_after":"86400","admob.app_id":"ca-app-pub-8529835372050931~4358271409","admob.unit.article_detail_bottom":"ca-app-pub-8529835372050931/1883637630","admob.unit.article_detail_top":"ca-app-pub-8529835372050931/8604898230","admob.unit.event_article_top":"ca-app-pub-8529835372050931/7406887709","admob.unit.interestial":"ca-app-pub-8529835372050931/2741937409","ios_appstore_link":"https://itunes.apple.com/us/app/loigiaihay/id1209891610?mt=8","android_appstore_link":"https://play.google.com/store/apps/details?id=com.tp.loigiaihay","min_version":"2.1.3","latest_version":"1.3.9","force_update":false},"popup_ads":[]}