حل التيار والمقاومة في فيزياء بالإجابات والشرح

1. AIncrease the cross-sectional area of the conductor to a double.

   ليه دي غلطDoubling the area halves the resistance, not quarters it.

2. BDecrease the diameter of the conductor to a half.

   ليه دي غلطHalving the diameter quarters the area, so the resistance becomes four times larger.

3. CIncrease the diameter of the conductor to a double. الإجابة الصحيحة
4. DDecrease the cross-sectional area of the conductor to a quarter.

   ليه دي غلطQuartering the area makes resistance four times larger, not one quarter.

ليه دي الصحFor the same material, length, and temperature, R = ρL/A, so resistance is inversely proportional to cross-sectional area. To make R become R/4, the area must become 4A. Since A ∝ d², doubling the diameter makes the area four times larger.

1. ABending the wire in the middle

   ليه دي غلطBending the wire does not make the whole wire four times as resistant in this model.

2. BBending the wire at one-third of its length

   ليه دي غلطBending at one-third of the length is not a uniform stretch that gives R′ = 4R.

3. CStretching the wire regularly to double its original length الإجابة الصحيحة
4. DStretching the wire regularly to reduce its diameter to half

   ليه دي غلطReducing diameter to half makes area one quarter; under uniform stretching, the resistance would increase much more than 4R.

ليه دي الصح"Increasing its resistance by 3R" means the final resistance is R + 3R = 4R. When the same wire is stretched uniformly to double its length, its volume stays constant, so its cross-sectional area becomes A/2. Therefore R′ = ρ(2L)/(A/2) = 4R.

1. A80 Ω

   ليه دي غلط80 Ω is the final resistance, not the increase in resistance.

2. B20 Ω

   ليه دي غلط20 Ω would come from only quartering the area and ignoring the length change.

3. C75 Ω الإجابة الصحيحة
4. D15 Ω

   ليه دي غلط15 Ω does not account correctly for area and length changes.

ليه دي الصحReducing the diameter to half makes the cross-sectional area one quarter of the original area. For the same wire reshaped uniformly, volume stays constant, so the length becomes four times the original length. Since R = ρL/A, the new resistance is 16R = 16 × 5 = 80 Ω. The increase is 80 − 5 = 75 Ω.

1. A2.25 A

   ليه دي غلط2.25 A comes from an incorrect equivalent resistance after reshaping the wire parts.

2. B3.75 A

   ليه دي غلط3.75 A is close, but it misses the exact parallel reduction of X and Z.

3. C4 A الإجابة الصحيحة
4. D1.2 A

   ليه دي غلط1.2 A treats the reshaped parts as much larger resistance than they are.

ليه دي الصحEach equal part starts with resistance 3 Ω. Stretching X to double its length keeps volume constant, so its area halves and RX becomes 4 × 3 = 12 Ω. Reshaping Y to double its area makes its length half, so RY becomes 3/4 = 0.75 Ω. Z remains 3 Ω. The parallel combination of X and Z is 2.4 Ω, so the total resistance is 3.15 Ω. Therefore I = 12.6/3.15 = 4 A.

1. APositive charge flows out from the negative terminal of the battery.

   ليه دي غلطOutside the battery, conventional current leaves the positive terminal, not the negative terminal.

2. BNegative charge flows out from the positive terminal of the battery.

   ليه دي غلطThis describes electron flow, not conventional current.

3. CPositive charge flows from the negative terminal to the positive terminal inside the battery. الإجابة الصحيحة
4. DPositive charge flows from the positive terminal to the negative terminal inside the battery.

   ليه دي غلطInside the battery, positive charge is driven from lower to higher potential, not from positive to negative.

ليه دي الصحConventional current is defined as the direction of positive charge flow. In the external circuit it leaves the positive terminal, while inside the battery the source moves positive charge from the negative terminal to the positive terminal to maintain the potential difference.

1. Athe length of conductor Y

   ليه دي غلطEqual lengths would make X have twice the resistance of Y.

2. Bhalf the length of conductor Y الإجابة الصحيحة
3. Ctwice the length of conductor Y

   ليه دي غلطTwice the length would make X four times the resistance of Y.

4. Dquarter the length of conductor Y

   ليه دي غلطA quarter length would make X half the resistance of Y.

ليه دي الصحResistance is R = ρL/A. Since dX = dY/2, the area of X is AY/4. Also ρX = ρY/2. Thus RX = (ρY/2)LX/(AY/4) = 2ρYLX/AY. For RX = RY = ρYLY/AY, we need 2LX = LY, so LX = LY/2.

1. AThe resistances of wires A, B, and C are equal.

   ليه دي غلطThe radii are different, so the resistances are not equal.

2. BThe resistance of wire C is half that of wire B.

   ليه دي غلطRC is two-thirds of RB, not half of RB.

3. CThe current through wire B is double the current through wire A.

   ليه دي غلطSeries components carry the same current.

4. DThe potential difference across wire A is three times that across wire C. الإجابة الصحيحة

ليه دي الصحFor wires of the same material, R ∝ L/r². Thus RA ∝ L/r², RB ∝ 2L/(2r)² = RA/2, and RC ∝ 3L/(3r)² = RA/3. In series, the same current passes through all wires, so voltage is proportional to resistance. Therefore VA = 3VC.

1. أتزيد

   ليه دي غلطإضافة مقاومة على التوازي بتفتح مسار إضافي للتيار، فالمقاومة الكلية بتقل مش بتزيد.

2. بتقل الإجابة الصحيحة
3. جثابتة

   ليه دي غلطالمقاومة الكلية بتتغير فعلاً لما تضيف مقاومة موازية، فهي مش ثابتة.

ليه دي الصحطريق تاني اتفتح — المرور بقى أسهل.

عند إضافة مقاومة على التوازي يظهر مسار جديد للتيار. في التوازي لا نجمع المقاومات مباشرة، بل نجمع مقاليبها: 1/R_total = 1/R1 + 1/R2 + ... لذلك يزداد مجموع المقاليب فتقل المقاومة الكلية. لهذا تكون المقاومة المكافئة بعد إضافة فرع موازٍ أقل من قبل، بل أقل من أصغر مقاومة منفردة في الفروع.