פתרון בגרות חשמל — שאלון 36371

|q₁| = |q₂| = 10⁻⁸ C | d = 6 cm = 0.06 m
k = 9·10⁹ N·m²/C²

1.0 cm

3.0 cm

2.00

0.71

8.9

0

−1.5

r = d2 = 0.062 = 0.03 m E₁ = k·|q|r² = 9·10⁹ · 10⁻⁸0.03² = 909·10⁻⁴ = 1·10⁵ V/m

E_A = E₁ + E₂ = 1·10⁵ + 1·10⁵ = 2·10⁵ V/m

V = k·qr ⟸

V_A = k·(+10⁻⁸)0.03 + k·(−10⁻⁸)0.03 = 3000 − 3000 = 0 V

F = q·E , q < 0 ⟹ F

E₁(C)E₁(A) = r_A²r_C² = 0.03²0.01² = 9

U = k·q₁·q₂d

U = 9·10⁹ · (+10⁻⁸) · (−10⁻⁸)0.06 = −9·10⁻⁷0.06

U = −1.5·10⁻⁵ J

ε = 90 V | r = 50 Ω | R₁ = 1000 Ω | L₂ = 0.75·L₁

45 °C

אידיאלי

0.050

37.5

1000

1750

3.00

R = ρ·LA

R₂R₁ = L₂L₁ = 0.75 R₂ = 0.75 · 1000 = 750 Ω

R_ext = R₁ + R₂ = 1000 + 750 = 1750 Ω

I = εr + R_ext = 9050 + 1750 = 901800 = 0.05 A

V_R₂ = I·R₂ = 0.05 · 750 = 37.5 V

I·r + I·R₁ + I·R₂ = 2.5 + 50 + 37.5 = 90 V = ε

I = qΔt ⟹ q = I·Δt

q = 0.05 · 60 = 3 C

I = V_R₂R₂ = 52750 ≈ 0.0693 A

ε = I·(r + R_th + R₂) 90 = 0.0693 · (50 + R_th + 750) 50 + R_th + 750 = 900.0693 ≈ 1298 Ω

R_th = 1298 − 800 ≈ 498 Ω ≈ 0.5 kΩ

1R_|| = 1R₂ + 1R_V ⟹ R_|| < R₂

R_tot = r + R₁ + R_|| < 1800 Ω

I = εR_tot

ε = 36 V | r = 6 Ω | R₁ = 12 Ω | Δt = 200 s

2.00

24.0

48.0

66.7

9600

I = εr + R₁ = 366 + 12 = 3618 = 2 A

P_R₁ = I²·R₁ = 2² · 12 = 48 W

E = P·Δt = 48 · 200 = 9600 J

η = P_extP_tot = I·V_tI·ε = V_tε

V_t = ε − I·r = 36 − 2·6 = 24 V

η = 2436 = 23 ≈ 66.7%

P_tot = ε·I

P_r = I²·r

P = ε·I − I²·r = I·(ε − I·r)

I·(36 − 6I) = 48

36I − 6I² = 48 ⟹ 6I² − 36I + 48 = 0 ⟹ I² − 6I + 8 = 0 I = 6 ± √36 − 322 = 6 ± 22 ⟹ I = 2 I = 4

R_ext = εI − r = 364 − 6 = 3 Ω 13 = 112 + 1R₂ ⟹ R₂ = 4 Ω

V_t = ε − I·r = 36 − 4·6 = 12 V η = 1236 = 13 ≈ 33.3%

η = P_extε·I

C²·Ωs² = Cs² · Ω = A²·Ω = P = W

I = 15 A | ℓ = 4 cm = 0.04 m | g = 10 m/s²
β = 0 , 36 , 48 , 72 , 90 , 120 · cos β = 1 , 0.81 , 0.67 , 0.31 , 0 , −0.5
F = 0.88 , 0.80 , 0.70 , 0.56 , 0.40 , 0.18 N

0°

1.00

0.48

0.88

0.80

40

F_{(magnet on rod)} = −F_{(rod on magnet)}

F = I·ℓ × B , ℓ , F ⟹ B : Y → X

F_mag = B·I·ℓ·sin(90° − β) = B·I·ℓ·cos β

F − m·g − B·I·ℓ·cos β = 0

F = m·g + B·I·ℓ·cos β

slope = ΔFΔ(cos β) = 0.88 − 0.401 − 0 = 0.48 N

slope = B·I·ℓ ⟹ B = slopeI·ℓ

B = 0.4815 · 0.04 = 0.480.6 = 0.8 T

cos β = 0 ⟹ F = m·g = 0.40 N

m = 0.4010 = 0.04 kg = 40 g

R = 1000 Ω | q_max = 0.004 C | τ = R·C

0.50 s

2.52

2.94

5.06

0.50

500

slope = ΔqΔt

i(t) = dqdt

ε = V_R + V_C = i·R + qC

i = ε − V_CR

q_max = ε·C·(1 − 0) = ε·C

q(τ) = ε·C·(1 − e^−τ/τ) = ε·C·(1 − e⁻¹)

q(τ)q_max = ε·C·(1 − e⁻¹)ε·C = 1 − e⁻¹ e⁻¹ ≈ 0.368 ⟹ 1 − 0.368 = 0.632 ≈ 63%

q_max = 0.004 C

0.63 · 0.004 = 0.00252 C ≈ 0.0025 C

τ = R·C ⟹ C = τR

C = 0.51000 = 5·10⁻⁴ F = 500 μF

Ω·F = VA·CV = CA = s

q_max = ε·C ⟹ ε = q_maxC

ε = 0.0045·10⁻⁴ = 8 V

A: t = 0.15 s , V = −5 mV | B: V = 0 | C: V ≈ +6 mV
משבצת אחת בגרף = 0.025 s · 1 mV = 2.5·10⁻⁵ V·s

0.150 s

−4.95

1.50

−1.6

1.6

0.0

Φ = B·A·cos θ

ε = − dΦdt

V_R = i·R

ε = − dΦdt ⟹

v = g·t ⟹ v_out > v_in

dΦdt = dΦdz · dzdt = dΦdz · v

ε = − ΔΦΔt ⟹ ΔΦ = −ε·Δt

[Φ] = [ε]·[t] = V·s

Wb = T·m² = N·mA = JA = JC·s = V·s

ΔΦ = −∫ ε·dt ⟹ |ΔΦ| =

1 = 0.025 · 0.001 = 2.5·10⁻⁵ V·s

|ΔΦ| ≈ 6 · 2.5·10⁻⁵ = 1.5·10⁻⁴ V·s

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