Physics Secondary الفيزياء 3

The internal resistance (r)

The internal resistance of a battery (electric cell) is a measure of the extent to which the battery itself hinders the flow of the electric current. The internal resistance depends on the size of the battery and on the material from which the battery is made. In some batteries, such as car batteries, it can be as low as 0.1 Ω; in others it can be higher than 100 Ω. The lower the internal resistance of a battery, the greater the intensity of the electric current it generates, and vice versa.

المقاومة الداخلية للبطارية (الخلية الكهربائية) هي مقياس لمدى إعاقة ( مقاومة ) البطارية نفسها لتدفق التيار الكهربائي. تعتمد المقاومة الداخلية على حجم البطارية وعلى المادة التي صنعت منها البطارية. في بعض البطاريات ، مثل بطاريات السيارات ، يمكن أن تصل إلى 0.1 Ω ؛ في حالات أخرى يمكن أن يكون أعلى من Ω100. كلما انخفضت المقاومة الداخلية للبطارية ، زادت شدة التيار الكهربائي الذي تولده ، والعكس صحيح.

The electromotive force (emf) of a battery and the potential difference between its terminals

From the definition of the electromotive force, if the electromotive force of a battery is denoted by (E), the total current intensity of a circuit is denoted by (I), the external resistance is denoted by (R), and the internal resistance of the battery is denoted by (r), we get:

القوة الدافعة الكهربائية (emf) البطارية وفرق الجهد بين أطرافها

من تعريف القوة الدافعة الكهربائية ، إذا تم الإشارة إلى القوة الدافعة الكهربائية للبطارية بواسطة (E) ، يتم الإشارة إلى شدة التيار الكلي للدائرة بالرمز (I) ، والمقاومة الخارجية يرمز إليها (R) ، والمقاومة الداخلية يتم الإشارة إلى البطارية بواسطة (r) ، نحصل على:

E = IR + Ir → (1) E = I (R + r) → (2) Therefore : I = ER + r

This relationship is known as Ohm's law for closed circuits, where:

The current intensity flowing through a circuit = Electromotive force Total resistance of the circuit

تُعرف هذه العلاقة بقانون أوم للدوائر المغلقة ، حيث: شدة التيار المتدفق عبر دائرة = القوة الدافعة الكهربائية المقاومة الكلية للدائرة

The relationship can be written as follows: يمكن كتابة العلاقة على النحو التالي:

∵ V = IR ∴ E = V + Ir

Where (V) is the potential difference between the ends of the external resistance, or the terminal potential difference, and can be measured by a voltmeter.

V = E - Ir

From this relationship, we find that the terminal potential difference (V) increases as the current intensity decreases. Note that the current intensity will decrease as the external resistance (R) in the circuit connected to the battery increases, see (Figure 1).

Example (1)

85 Ω, 70 Ω and 25 Ω are the respective resistances of three resistors connected in series to a battery of an electromotive force of 45 V. If the internal resistance of the battery is very small and can be omitted from the relevant formulas, calculate:

1. The current intensity flowing across the three resistors.

2. The potential difference applied on each resistor.

Solution:

The equivalent resistance of the circuit is equal to:

• R = R₁ + R₂ + R₃ = 85 + 70 + 25 = 180 Ω

• The total current intensity can be determined by Ohm’s law for closed circuits (omitting the internal resistance of the battery):

I = ER = 45180 = 0.25 A

Since the three resistors are connected in series, the current passing across each of them is the same 0.25 A.

• The potential difference across the first resistor is:
  V₁ = IR₁ = 0.25 × 85 = 21.25 V
  The potential difference across the second resistor is:
  V₂ = IR₂ = 0.25 × 70 = 17.5 V

• The potential difference across the third resistor is:
  V₃ = IR₃ = 0.25 × 25 = 6.25 V

Example (2)

If the three resistors in (example 1) are connected in parallel to the same source, calculate:

1. The current intensity flowing through each resistor.

2. The equivalent resistance.

3. The total current intensity.

Solution:

Since the three resistors are connected in parallel, the potential difference across each resistor (when the internal resistance of the battery is neglected) is 45 V. The current intensity for each resistor can be calculated by the following equations:

I₁ = VR₁ = 4585 = 0.529 A

I₂ = VR₂ = 4570 = 0.643 A

I₃ = VR₃ = 4525 = 1.8 A

The equivalent resistance is equal to:

1R_eq = 1R₁ + 1R₂ + 1R₃ = 185 + 170 + 125 R_eq = 15.14 Ω

The total current intensity can be determined by the following equation:

I_T = VR_eq = 4515.14 = 2.972 A

The total current intensity can also be determined by adding I₁, I₂ and I₃:

I_T = 1.8 + 0.643 + 0.529 = 2.972 A

Example (3)

A and B are two resistors, where A is connected in parallel to B, and both of them are connected in series to a resistor C and a battery of emf = 18 V. If A, B and C have resistances of 3 Ω, 6 Ω and 7 Ω respectively, and the internal resistance of the battery can be neglected, calculate:

1. The equivalent resistance.

2. The intensity of the current passing through the circuit.

3. The intensity of the current passing through each of A and B.

Solution:

The resistance equivalent to A and B, which are connected in parallel, is found by the following equation:

R` = R_A R_BR_A + R_B = 3 × 63 + 6 = 2 Ω

The resistance equivalent to A, B and C is calculated by the following equation:

R = R` + R_C = 2 + 7 = 9 Ω

The total current intensity is determined by the following equation:

I_T = ER = 189 = 2 A

To calculate the intensity of the current flowing through each of A and B, first calculate the potential difference:

V = IR` = 2 × 2 = 4 V

∴ I_A = VR_A = 43 = 1.333 A

I_B = VR_B = 46 = 0.667 A

Example (4)

An electric cell with an emf of 2 V is connected to a circuit. If the internal resistance of the cell is 0.1 Ω, and the external resistance of the circuit is 3.9 Ω, calculate the total current intensity flowing through the circuit.

Solution:  

   I = ER + r = 23.9 + 0.1 = 0.5 A  

Kirchhoff's voltage law

This law is based on the conservation of energy law and uses the insight that the electromotive force of a closed circuit represents the energy or work required for charges to flow through a circuit. The law states that around any closed loop (starting and ending at the same point) in a circuit, the sum of the electromotive forces is equal to the sum of the potential differences.

قانون الجهد كيرشوف

يعتمد هذا القانون على قانون الحفاظ على الطاقة ويستخدم البصيرة القائلة بأن القوة الدافعة الكهربائية لدائرة مغلقة تمثل الطاقة أو العمل المطلوب لتدفق الشحنات عبر الدائرة. ينص القانون على أنه حول أي حلقة مغلقة (تبدأ وتنتهي عند نفس النقطة) في دائرة ما ، فإن مجموع القوى الدافعة الكهربائية يساوي مجموع الفروق المحتملة.

If Kirchhoff’s voltage law is applied on we get that the algebraic sum of electromotive forces = V_s

The algebraic sum of the potential differences around the loop = IR₁+ IR₂

According to Kirchhoff’s voltage law:

The algebraic sum of electromotive forces = the algebraic sum of the potential differences around a closed loop

V_s = IR₁ + IR₂ ∴ V_s - IR₁ - IR₂ = 0

From this equation, we can derive another statement for Kirchhoff's voltage law, which is the algebraic sum of potential differences around any closed loop is zero. The direction of the current must be considered; the potential difference is negative when the current traverse a resistor in the positive direction, and vice versa. As for the electromotive force, it is negative when the current flows in the negative direction, and vice versa.

من هذه المعادلة ، يمكننا اشتقاق بيان آخر لقانون الجهد في كيرشوف، وهو المجموع الجبري للاختلافات المحتملة حول أي حلقة مغلقة هو صفر. يجب مراعاة اتجاه التيار ؛ يكون فرق الجهد سالبًا عندما يجتاز التيار المقاومة في الاتجاه الموجب والعكس صحيح. أما القوة الدافعة الكهربائية فهي سالبة عندما يتدفق التيار في الاتجاه السالب والعكس صحيح.

Therefore, the mathematical formula of Kirchhoff's voltage law is: لذلك ، فإن الصيغة الرياضية لقانون جهد كيرشوف هي:

∑ V = ∑ IR

Steps for solving problems by applying Kirchhoff's laws خطوات حل المشكلات بتطبيق قوانين كيرشوف

To find the current intensity and potential difference of a complex circuit, follow the steps below:

للبحث عن شدة التيار وفرق الجهد المحتمل لدائرة معقدة ، اتبع الخطوات التالية

Write the equation of Kirchhoff's current law, taking into consideration the direction of the current; it is considered negative if it is entering a node, and positive if it is leaving.

1. Assume the direction of the closed loop, and let it be the positive direction.

2. Apply Kirchhoff's voltage law to the closed loop, and write the equation.

3. The equation has two unknowns, which you can find by cancelling or substitution.

4. Now you can calculate any current intensity and through Ohm's law, you can find the potential difference applied on any resistor. If the current intensity is found to be negative, the real direction of the current is opposite to the direction assumed.

5. اكتب معادلة قانون كيرشوف الحالي ، مع مراعاة اتجاه التيار ؛ تعتبر سالبة إذا كانت تدخل عقدة ، و موجبة إذا كانت تخرج من الحلقة

6. افترض اتجاه الحلقة المغلقة ، واجعلها الاتجاه الإيجابي. _ طبق قانون جهد كيرشوف على الحلقة المغلقة ، واكتب المعادلة. _ تحتوي المعادلة على مجهولين ، يمكنك إيجادهما عن طريق الإلغاء أو الاستبدال.

7. يمكنك الآن حساب أي شدة تيار ومن خلال قانون أوم ، يمكنك إيجاد فرق الجهد المطبق على أي مقاومة. إذا تبين أن شدة التيار سالبة ، فإن الاتجاه الحقيقي للتيار يكون عكس الاتجاه المفترض.

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EXample :

Physics - 3 Sec شرح فيزياء لغات للثانوية العامة العبقرى - The Genius

https://www.youtube.com/playlist?list=PLOG5Av5r6CFiMogjWWFBdRl8hfZmb0xU2

مراجعة ليلة الامتحان Final Revision

Lecture 32\\ Final revision (Chapters 3&4) || 3rd sec