פתרון בגרות חשמל — שאלון 36371

R = 30 cm | V(R) = 90,000 V | V(∞) = 0 | OP = 12 cm
m₁ = 4·10⁻⁴ kg | q₁ = −1.2·10⁻⁸ C | m₂ = m₁ | q₂ = −q₁ | v = 1.5 m/s

54.0

50.0

92.6

1.50

4.50

V = kQr kQ = V(R) · R = 90000 · 0.30 = 27000 V·m Q = 270009·10⁹ = 3·10⁻⁶ C

AO = kQV_A = 2700050000 = 0.54 m

E = kQr² = 270000.54² = 270000.2916 = 9.26·10⁴ V/m

E = V_Ar = 500000.54 = 9.26·10⁴ V/m

OP = 12 cm < R = 30 cm

E(P) = 0

V(P) = V(R) = 90,000 V

½m₁v_A² + q₁V_A = ½m₁v_B² + q₁V_B

½ · 4·10⁻⁴ · 1.5² = 4.5·10⁻⁴ J q₁V_A = (−1.2·10⁻⁸) · 50000 = −6·10⁻⁴ J E_כוללת = 4.5·10⁻⁴ − 6·10⁻⁴ = −1.5·10⁻⁴ J

q₁V_B = (−1.2·10⁻⁸) · 40000 = −4.8·10⁻⁴ J E_k(B) = −1.5·10⁻⁴ − (−4.8·10⁻⁴) = 3.3·10⁻⁴ J

v_B = √2E_k(B)m₁ = √2 · 3.3·10⁻⁴4·10⁻⁴ = √1.65 = 1.28 m/s

q₂V_A = (+1.2·10⁻⁸) · 50000 = +6·10⁻⁴ J E_כוללת = 4.5·10⁻⁴ + 6·10⁻⁴ = 1.05·10⁻³ J

E_k(C) = 3.3·10⁻⁴ J q₂V_C = 1.05·10⁻³ − 3.3·10⁻⁴ = 7.2·10⁻⁴ J V_C = 7.2·10⁻⁴1.2·10⁻⁸ = 60,000 V

r_C = 2700060000 = 0.45 m , r_B = 2700040000 = 0.675 m AC = 54 − 45 = 9 cm , AB = 67.5 − 54 = 13.5 cm

הזרם נכנס אל הנגד המשתנה דרך המגע הנייד P ויוצא ב-K | הקטע MP הוא קצה מת
ההתנגדות הפעילה במעגל היא R(PK) בלבד

0.0

2.00

16.0

0.0

V₁ = ε − I·r

V₂ = I·R

שיפוע = 16 − 222.0 − 0.5 = −61.5 = −4 Ω r = 4 Ω

ε = V₁ + I·r = 22 + 0.5 · 4 = 24 V ε = 16 + 2.0 · 4 = 24 V

R = V₂I = 16 − 42.0 − 0.5 = 121.5 = 8 Ω R = 40.5 = 8 Ω , R = 162.0 = 8 Ω

I_min = 0.5 A

R_total = εI_min = 240.5 = 48 Ω

R_PK = 48 − r − R = 48 − 4 − 8 = 36 Ω

I = 244 + 8 = 2.0 A , V₁ = 24 − 2.0 · 4 = 16 V

V_PK = ε − I(r + R) = 24 − 12·I

R_PK: 0 → 36 Ω ⇒ I: 2.0 → 0.5 A ⇒ V_PK: 0 → 18 V

I = εr + R + R_PK + ∞ = 0

V₂ = I·R = 0 · 8 = 0

V₁ = ε − I(r + R + R_PK) = ε = 24 V

ε = 24 V | L₁ · 24 V · 20 W | L₂ · 12 V · ההספק נמחק
הגרף של V₁ כפונקציה של V₂ הוא ישר העובר דרך הראשית, ושיפועו 4.5

0.0

0.833

24.0

5.33

4.50

20.0

P = V²R → R_L1 = V²P = 24²20 = 57620 = 28.8 Ω

I = 2428.8 = 0.833 A , P = 24 · 0.833 = 20 W

I_L1 = PV = 2024 = 0.833 A

I_R = I_L2 = I_L1 = 0.833 A

V_מקבילי = 24 V = ε

V_PK = ε − V_מקבילי = 24 − 24 = 0 R_PK = 0I = 0

V₁ = I · R_L1 , V₂ = I · R_L2

I = V₂R_L2 → V₁ = R_L1R_L2 · V₂

שיפוע = 24.3 − 05.4 − 0 = 4.5

R_L1R_L2 = 4.5 → R_L2 = 28.84.5 = 6.4 Ω

P_L2 = V²R_L2 = 12²6.4 = 1446.4 = 22.5 W

I_L1 = I_L2 = 0.833 A , I_כולל = 2 · 0.833 = 1.667 A

V_L2 = 0.833 · 6.4 = 5.33 V , V_R = 24 − 5.33 = 18.67 V R = 18.670.833 = 22.4 Ω

P_L1 = 20 W P_L2 = I²R_L2 = 0.833² · 6.4 = 4.44 W P_R = I²R = 0.833² · 22.4 = 15.56 W P_רכיבים = 20 + 4.44 + 15.56 = 40.0 W

η = P_רכיביםP_מקור = P_רכיביםε₁ · I_כולל = 0.8 ε₁ = 40.00.8 · 1.667 = 40.01.333 = 30 V

x (m) 0.10 · 0.15 · 0.20 · 0.25 · 0.30
α (°) 29 · 20 · 16 · 13 · 9
tan α 0.55 · 0.36 · 0.29 · 0.23 · 0.16

0.10 m

0.0 A

0.10

10.0

29

0.55

16.0

tan α = B_תילB_E∥

B_תיל = μ₀I₁2πx

tan α = B_תילB_E∥ = μ₀I₁2πx · B_E∥

tan α = μ₀I₁2π B_E∥ · 1x

1x [m⁻¹]

x (m) 0.10 0.15 0.20 0.25 0.30 1 חלקי x (m⁻¹) 10.0 6.67 5.00 4.00 3.33 tan α 0.55 0.36 0.29 0.23 0.16

שיפוע = 0.5510.0 ≈ 0.055 m

שיפוע = μ₀I₁2π B_E∥ → B_E∥ = μ₀I₁2π · שיפוע

μ₀2π = 2·10⁻⁷ T·m/A B_E∥ = 2·10⁻⁷ · 80.055 = 1.6·10⁻⁶0.055 = 2.9·10⁻⁵ T

B₁ + B₂ = −B₁ → B₂ = −2B₁

B₂B₁ = I₂I₁ = 2 → I₂ = 2I₁ = 2 · 5 = 10 A

v = 10⁶ m/s | B₁ = 0.12 T | ΔV = 800 V | Δx = 5 cm
m_p = 1.673·10⁻²⁷ kg | q_p = 1.602·10⁻¹⁹ C | כיוון B₂ זהה לכיוון B₁

0.0160 T

8.70

16.0

0.0160

1.0000

140.0

F = q · v × B

qvB₁ = mv²r → r = mvqB₁ r = 1.673·10⁻²⁷ · 10⁶1.602·10⁻¹⁹ · 0.12 = 1.673·10⁻²¹1.922·10⁻²⁰ = 8.7·10⁻² m

E = ΔVΔx = 8000.05 = 16,000 V/m = 1.6·10⁴ V/m

qE + q · v × B₂ = 0

qE = qvB₂ → B₂ = Ev

B₂ = 1.6·10⁴10⁶ = 1.6·10⁻² T

W = qE · d > 0

W = ΔE_k > 0 → v_סופית > v

F_∥ = 0 → v_∥ = v = const

t = d_∥v → t_{ללא B₂} = t_{עם B₂}

a = 50 cm | A = 0.25 m² | R = 2 Ω
פרק II · B(t) = −t² + 1.2t − 0.11 · שיא של 0.25 T ברגע t = 0.6 s

0.00 s

0.00

0.090

0.0225

0.000

A = a² = 0.50² = 0.25 m² Φ = B·A , ε = −ΔΦΔt , I = εR

Φ = 0.09 · 0.25 = 0.0225 Wb ε = 0 , I = 0

Φ = 0.25 · (−t² + 1.2t − 0.11) = −0.25t² + 0.3t − 0.0275 Wb ε = −dΦdt = −(−0.5t + 0.3) = 0.5t − 0.3 V I = εR = 0.5t − 0.32 = 0.25t − 0.15 A

Φ = 0.0225 Wb , ε = 0 , I = 0

B(0.2) = −0.04 + 0.24 − 0.11 = 0.09 T , B(1.0) = −1 + 1.2 − 0.11 = 0.09 T B(0.6) = −0.36 + 0.72 − 0.11 = 0.25 T , Φ_max = 0.0625 Wb

B(0.3) = −0.09 + 0.36 − 0.11 = 0.16 T > B(0.2) = 0.09 T

I(0.3) = 0.25 · 0.3 − 0.15 = −0.075 A

I(0.2) = −0.10 A , I(0.6) = 0 , I(1.0) = +0.10 A

I_max = 0.25 · 1.0 − 0.15 = 0.10 A

q = I·Δt = εR·Δt = 1R·ΔΦΔt·Δt = ΔΦR

Φ(0) = 0.0225 Wb , Φ(0.6) = 0.25 · 0.25 = 0.0625 Wb q = ΔΦR = 0.0625 − 0.02252 = 0.042 = 0.02 C

Φ(1.2) = 0.0225 Wb → ΔΦ = 0.0225 − 0.0625 = −0.04 Wb q = 0.042 = 0.02 C

פתרון בגרות · חשמל