פתרון בגרות חשמל — שאלון 36371

σ [10⁻⁷ C/m²] · 1.50 2.25 3.25 4.00 5.00
α [°] · 4 6 8 10 12
tan(α) · 0.07 0.11 0.14 0.18 0.21
m₁ = 1 gr | g = 10 m/s² | ε₀ = 8.85·10⁻¹² C²/N·m²

3.25 ·10⁻⁷ C/m²

8.0

0.14

18

1.4

4.4

m₁·g = 10⁻³ · 10 = 10⁻² N

T ⟸

F_E = |q₁|·E , E = σ2·ε₀

(1.50 , 0.07) (2.25 , 0.11) (3.25 , 0.14) (4.00 , 0.18) (5.00 , 0.21)

σ = 0 ⟹ E = 0 ⟹ F_E = 0 ⟹ α = 0 ⟹ tan(α) = 0

slope = Δ tan(α)Δ σ = tan(α) − 0σ − 0

slope ≈ 0.2195.00 = 0.0438 [10⁻⁷ C/m²]⁻¹

slope = 0.043810⁻⁷ = 4.4·10⁵ m²/C

T·cos(α) = m₁·g | T·sin(α) = |q₁|·E

tan(α) = T·sin(α)T·cos(α) = |q₁|·Em₁·g

E = σ2·ε₀

tan(α) = |q₁|m₁·g · σ2·ε₀ = |q₁|2·ε₀·m₁·g · σ

F_E = q₁·E , F_E ↔ E ⟹ q₁ < 0

slope = |q₁|2·ε₀·m₁·g ⟹ |q₁| = slope · 2·ε₀·m₁·g

|q₁| = 4.38·10⁵ · 2 · 8.85·10⁻¹² · 10⁻³ · 10 = 7.7·10⁻⁸ C

F_לוח = |q|·σ2·ε₀ ⟸

ε = 6 V | r = 1 Ω | R_MN(max) = 12 Ω
R₁ = 2 Ω | R₂ = 3 Ω

N · 12.0 Ω

3.0 V

0.33

5.67

0

0.67

4.00

R_tot = r + R_MN + R₁ + R₂ = 1 + 12 + 2 + 3 = 18 Ω

I = εR_tot = 618 = 13 ≈ 0.33 A

V(S₂) = I·R₁ = 13 · 2 = 23 ≈ 0.67 V

I = εr + R_M→גררה + R₁ + R₂ ⟸

V = ε − I·r = 6 − I·1

I = 61 + 12 + 3 = 616 = 0.375 A

I(R₁) = 0 ⟹ V(R₁) = 0

V(S₁) = V(R₁ ⟶) − V() = V() − V() = 0

R_MN = R_M→גררה + R_גררה→N = 12 Ω ⟸

I = 61 + 12 + 3 = 0.375 A , V = 6 − 0.375 · 1 = 5.625 V

V(M) = V(N) = V()

ε = I·r + I·R₂ ⟹ 6 = I·(1 + 3) = I·4 I = 64 = 1.5 A

ε* = I·R₁ = 1.5 · 2 = 3 V

V(D) − V(C) = +3 V ⟹ D

L₁ : 18 V , 27 W
: ε₁ = 30 V , r₁ = 2 Ω | : ε₂ = 32 V , r₂ = 10 Ω

6.0 Ω

1

1.50

18.0

27

45

60

P = V·I ⟹ I = PV = 2718 = 1.5 A

R_L = VI = 181.5 = 12 Ω

I·r₂ = 1.5 · 10 = 15 V

V_term = ε₂ − I·r₂ = 32 − 15 = 17 V

I = 3210 + 12 = 1.45 A , V_L = 1.45 · 12 = 17.5 V < 18 V

ε₁ = I·(r₁ + R_L + R_MN) 30 = 1.5 · (2 + 12 + R_MN) R_MN = 301.5 − 14 = 20 − 14 = 6 Ω

V_MN = 30 − 1.5·2 − 18 = 30 − 3 − 18 = 9 V R_MN = 91.5 = 6 Ω

R_par = R_Ln = 12n

I_tot = n · 1.5 A

ε₁ = I_tot·r₁ + V_L ⟹ 30 = 1.5n · 2 + 18

3n = 30 − 18 = 12 ⟹ n = 123 = 4

I_tot = 4 · 1.5 = 6 A , I·r₁ = 6 · 2 = 12 V , 30 − 12 = 18 V ✓

P_L = 27 W , P_tot = ε₁·I = 30 · 1.5 = 45 W η = 2745 = 0.60 = 60%

P_L = 4 · 27 = 108 W , P_tot = 30 · 6 = 180 W η = 108180 = 0.60 = 60%

η = V_L·Iε₁·I = V_Lε₁ = 1830 = 60%

a = c = 1 m | b = d = 30 cm | h = 10 cm
I₂ [A] · 0 3 6 9 12 15 18
N [N] · 0.0400 0.0403 0.0406 0.0409 0.0412 0.0415 0.0418

9.0 A

0.0409

1.20

0.30

0.90

0.0400

N = m·g + F_net ⟸

B = μ₀·I₁2·π·r

F_aF_c = B_aB_c = h + bh = 0.400.10 = 4

I₂ → : x̂ × (−ẑ) = +ŷ ⟹ I₂ ← : (−x̂) × (−ẑ) = −ŷ ⟹ ✓

(−ŷ) × (−ẑ) = +x̂ | (+ŷ) × (−ẑ) = −x̂

F = B·I₂·L = μ₀·I₁2·π·r · I₂ · a

F_a = μ₀·I₁·I₂·a2·π·h , F_c = μ₀·I₁·I₂·a2·π·(h + b)

F_net = F_a − F_c = μ₀·I₁·a2·π · 1h · I₂ − μ₀·I₁·a2·π · 1h + b · I₂

N = m·g + μ₀·I₁·a2·π · ( 1h − 1h + b ) · I₂

N(I₂ = 0) = m·g = 0.0400 N

m = 0.040010 = 4·10⁻³ kg = 4 gr

slope = 0.0418 − 0.040018 − 0 = 1.8·10⁻³18 = 1.0·10⁻⁴ N/A

1h − 1h + b = 10.10 − 10.40 = 10 − 2.5 = 7.5 m⁻¹

μ₀2·π = 4·π·10⁻⁷2·π = 2·10⁻⁷ T·m/A

1.0·10⁻⁴ = 2·10⁻⁷ · I₁ · 1 · 7.5 I₁ = 1.0·10⁻⁴1.5·10⁻⁶ = 66.7 A

ε = 24 V | R₁ = 100 Ω | R₂ = 140 Ω | C = 0.1 μF

0.0 μs

0.100

0.0

0.00

—

I = εR₁ + R₂ = 24100 + 140 = 24240 = 0.1 A

V(R₂) = I·R₂ = 0.1 · 140 = 14 V

I_C(∞) = 0 ⟹ I = 24240 = 0.1 A

V_C = V(R₂) = 14 V

Q = C·V_C = 0.1·10⁻⁶ · 14 = 1.4·10⁻⁶ C = 1.4 μC

I(0 < t < t₁) = εR₁ + R₂ = 0.1 A

I(t₁⁺) = εR₁ = 24100 = 0.24 A

I(t) = εR₁ + R₂ + ( εR₁ − εR₁ + R₂ )·e^{−(t−t₁)/τ}

Q(t) = 1.4·10⁻⁶ · e^−t/τ ⟶ 0

I(R₂) = V_C(t)R₂ = 14140·e^−t/τ = 0.1·e^−t/τ A ⟶ 0

τ_pf = R₂·C = 140 · 0.1·10⁻⁶ = 1.4·10⁻⁵ s = 14 μs

τ_tn = R₁·R₂R₁ + R₂·C = 100 · 140240·0.1·10⁻⁶ = 5.8 μs

= 9 m | ρ = 1.5·10⁻⁶ Ω·m | A_c = 5 mm²
B(t) = (6t − t²)·10⁻³ T · 0 < t ≤ 3 s

2.00 s

0°

10.8

8.00

2.00

0.162

15.0

L = 4 · 9 = 36 m

A_c = 5 mm² = 5·10⁻⁶ m²

R = ρ·LA_c = 1.5·10⁻⁶ · 365·10⁻⁶ = 54·10⁻⁶5·10⁻⁶ = 10.8 Ω

Φ = B(t)·A

ε = − ΔΦΔt = − A · ΔBΔt

dBdt = (6 − 2t)·10⁻³ T/s

t = 0.5 : dBdt = 5·10⁻³ ⟹ ε = 81 · 5·10⁻³ = 0.405 V ⟹ I = 0.40510.8 = 0.0375 A t = 2.5 : dBdt = 1·10⁻³ ⟹ ε = 81 · 1·10⁻³ = 0.081 V ⟹ I = 0.08110.8 = 0.0075 A

0 < t < 3 ⟹ dBdt = (6 − 2t)·10⁻³ > 0

A = 9² = 81 m²

dBdt = (6 − 2·2)·10⁻³ = 2·10⁻³ T/s

|ε| = A · dBdt = 81 · 2·10⁻³ = 0.162 V

I = |ε|R = 0.16210.8 = 0.015 A = 15 mA

Φ = B·A·cos(α) , 0 < cos(α) < 1

|ε| = A·cos(α)·dBdt < A·dBdt

I = A·cos(α)R·dBdt = 0.015 · cos(α) < 0.015 A

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