الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

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Direct Current CircuitsHints for solution of problems using Kirchhoff’s rulesلاستخدام قاعدة آيرشوف يجب اتباع الخطوات التالية:‏emf(1)(2)(3)(4)(5)حدد اتجاه القوة الدافعة الكهربية لكل بطارية في الدائرة الكهربية بسهم متجه منالقطب السالب إلى القطب الموجب للبطارية.‏حدد اتجاه التيار الكهربائي لكل عنصر من عناصر الدائرة الكهربية مثل المقاومةبافتراض اتجاه محدد للتيار حتى تتمكن من تطبيق قاعدتي آيرشوف.‏ فإذا آان الحلالنهائي يظهر إشارة موجبة للتيار يكون الاتجاه المفترض صحيحاً،‏ أما إذا ظهرت إشارةالتيار سالبة فإن قيمة التيار صحيحة،‏ ولكن اتجاه التيار في الاتجاه المعاآس للاتجاهالمفترض.‏نطبق القاعدة الأولى لكيرشوف عند العقدة الموجودة في الدائرة الكهربية بحيث تكونإشارة التيارات الداخلة على العقدة موجبة والخارجة من العقدة سالبة.‏نطبق القاعدة الثانية لكيرشوف على مسار مغلق محدد لكل فرع من أفرع الدائرة الكهربيةونراعي التغير في فرق الجهد على آل عنصر من عناصر الدائرة الكهربية إذا آان سالباً‏أو موجباً.‏نحل المعادلات الرياضية التي نتجت من تطبيق الخطوتين(3)و(‏‎4‎‏)‏ حلاً‏ جبرياً.‏www.hazemsakeek.com
Lectures in General Physics8.4 Single-Loop CircuitIn a single-loop circuit there is no junctions and the current is the same in allelements of the circuit, therefore we use only the second Kirchhoff rule.Example 8.3Two battery are connected in opposite in a circuit contains two resistorsas shown in figure 8.6 the emf and the resistance are ξ 1 =6v, ξ 2 =12v,R 1 = 8Ω, and R 2 =10 Ω. (a) Find the current in the circuit. (b) What isthe power dissipated in each resistor?abIIξ 1R 2Iξ 2IR 1dcFigure 8.6SolutionFrom figure 8.6 the circuit is a single-loop circuit. We draw an arrow foreach emf in the circuit directed from the -ve to +ve terminal of the emf. Ifwe assume the current in the circuit is in the clockwise direction (abcda).Applying the second Kirchhoff’s rule along arbitrary loop (abcda) we get∑ ∆Vii= 0Starting at point (a) to point (b) we find the direction of the loop is the sameas the direction of the emf therefore ξ 1 is +ve, and the direction of the loopfrom point (b) to point (c) is the same with the direction of the current thenthe change in potential difference is -ve and has the value -IR 1 . Completethe loop with the same principle as discussed before we get;Dr. Hazem Falah Sakeek
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LECTURES INGENERAL PHYSICSPart TwoE
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محاضرات في الفيزيا
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ContentsChapter (1) Electrostatic1.
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4.5 Energy stored in a charged capa
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Introduction to ElectrostaticsPart
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Introduction to ElectrostaticsIntro
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Introduction to Electrostaticssame
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Introduction to Electrostatics1.2 P
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Introduction to ElectrostaticsRubbe
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Coulomb’s Lawقانون آولو
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Coulomb’s LawCoulomb’s lawقا
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Coulomb’s Law2.2 Calculation of t
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Coulomb’s Lawq 1 &q 3نتيجة
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Coulomb’s Lawx∑∑FFxy= 2 × 0.
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Coulomb’s LawExample 2.4A charge
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Coulomb’s LawExample 2.5Two fixed
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Coulomb’s Law2.3 Problems2.1) Two
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Lectures in General Physicsالكه
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Lectures in General PhysicsElectric
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3.3 The direction of E rLectures in
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Lectures in General PhysicsExample
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Lectures in General PhysicsE y = E
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Lectures in General PhysicsE y = -E
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Electric Flux4.1 The Electric Flux
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Lectures in General PhysicsElectric
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6.3.2 Cylindrical capacitorLectures
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6.4.2 Capacitors in series:Lectures
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Lectures in General PhysicsAPPENDIX
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LengthMetric and Imperial Measures1
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الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

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