פתרון בגרות חשמל — שאלון 36371

r (m)      | 0.16 | 0.18 | 0.20 | 0.22 | 0.24
T_B (N)   | 0.41 | 0.33 | 0.28 | 0.25 | 0.21
k = 9·10⁹ N·m²/C²   ·   g = 10 m/s²

0.20 m

0.20

25.00

0.224

0.282

0.166

F_g = mg

T_B

F_E = k|q · (−q)|r² = kq²r²

T_B = mg + F_E = mg + kq²r²

T_B = kq² · 1r² + mg מול y = שיפוע · x + חיתוך

x = 1r² , [x] = m⁻²

r (m) | 0.16 | 0.18 | 0.20 | 0.22 | 0.24
T_B (N) | 0.41 | 0.33 | 0.28 | 0.25 | 0.21
1/r² (m⁻²) | 39.06 | 30.86 | 25.00 | 20.66 | 17.36

T_B = kq² · 1r² + mg

kq² = ΔT_BΔ(1/r²) = 8.96·10⁻³ N·m² mg = 5.77·10⁻² N

|q| = √kq²k = √8.96·10⁻³9·10⁹ = √9.96·10⁻¹³ = 1.0·10⁻⁶ C

m = mgg = 5.77·10⁻²10 = 5.77·10⁻³ kg

r (m)              | 0.16   | 0.18   | 0.20   | 0.22   | 0.24
T_B נמדד (N)  | 0.41   | 0.33   | 0.28   | 0.25   | 0.21
T_B מודל (N)  | 0.408 | 0.334 | 0.282 | 0.243 | 0.213

T_B = F_E + mg

F_E = T_A + mg → T_A = F_E − mg

T_B − T_A = (F_E + mg) − (F_E − mg) = 2mg > 0

T_B = 0.350 + 0.058 = 0.408 N , T_A = 0.350 − 0.058 = 0.292 N

ε = 20 V | r = 2 Ω | V_הדקים = 16 V
V₁ = 12 V | V₂ = 15 V (במיקום החדש של המגע P)

0.909

13.2 Ω

12.00

1.20

2.50

1.50

12.00

15.00

V_r = ε − V = 20 − 16 = 4 V

I = ε − Vr = 42 = 2 A

R₁ = VI = 162 = 8 Ω

V₂ = V₁ + V_PN

R_MP = R_PN = R_MN2

R_par = R₁ · R_MPR₁ + R_MP < R_MP = R_PN

V₁V_PN = R_parR_PN < 1 → V₁ < V₂2

I = ε − V₂r = 20 − 152 = 2.5 A

I₁ = V₁R₁ = 128 = 1.5 A

I_MP = I − I₁ = 2.5 − 1.5 = 1.0 A

R_MP = V₁I_MP = 121.0 = 12 Ω

V_PN = V₂ − V₁ = 15 − 12 = 3 V

R_PN = V_PNI = 32.5 = 1.2 Ω

R_MN = R_MP + R_PN = 12 + 1.2 = 13.2 Ω

R_par = 8 · 128 + 12 = 4.8 Ω R_ext = 4.8 + 1.2 = 6 Ω , I = 206 + 2 = 2.5 A V₂ = 2.5 · 6 = 15 V , V₁ = 2.5 · 4.8 = 12 V

R_ext = R_MN , I = εR_MN + r

V₂ = ε − I·r = קבוע , V₁ = I·R_MP גדל

V₁ = 0

V₂ = ε·R_PNR_PN + r קטן

I (A)  | 1.3 | 2.2 | 2.7 | 3.7 | 4.5
V (V)  | 14.1 | 11.4 | 9.9 | 6.9 | 4.5
P (W)  | 18.3 | 25.1 | 26.7 | 25.5 | 20.3

6.0 Ω

6.0

2.00

12.00

24.00

12.00

66.7

ε = V + I·r → V = ε − I·r V = שיפוע · I + חיתוך , שיפוע = −r , חיתוך = ε

שיפוע = ΔVΔI = 4.5 − 14.14.5 − 1.3 = −9.63.2 = −3 V/A r = 3 Ω

ε = V + I·r = 14.1 + 1.3·3 = 18 V ε = 4.5 + 4.5·3 = 18 V

P = V·I

V = ε − I·r → I = ε − Vr

P = V · ε − Vr = εV − V²r = −V²r + εVr

P = 18·9.9 − 9.9²3 = 178.2 − 98.013 = 26.7 W

Q = P·t → t = QP

V_max = ε2 = 182 = 9 V P_max = ε²4r = 18²4·3 = 32412 = 27 W

I = εR + r = 183 + 3 = 3 A , V = I·R = 9 V , P = I²R = 27 W R_QN = r = 3 Ω

P = I²R = ε²R(R + r)²

η = P_RP_total = I²RI²(R + r) = RR + r η = rr + r = 12 = 50%

1 eV = e · 1 V = 1.6·10⁻¹⁹ J P = Et → 1 eVs = 1.6·10⁻¹⁹ W

V = ε − I·r

I = εR_QN + R_A + r

מערכת 1 | R₁ = 0.15 m | N₁ = 10 | i₁ ניתן לשינוי בגודל ובכיוון
מערכת 2 | R₂ = 0.10 m | N₂ = 10 | i₂ קבוע בגודל ובכיוון

0.00

0.0

31.4

1.00

45

B = μ₀Ni2R

tan θ = B₂B_E∥ = 1 → θ = 45°

B₁ = μ₀N i2R₁ , B₂ = μ₀N i2R₂ , R₁ > R₂ ⇒ B₂ > B₁

B₂ − B₁ = 31.4 − 20.9 = 10.5 μT tan θ = 10.531.4 = 0.33 → θ ≈ 18°

μ₀N₁|i₁|2R₁ = μ₀N₂i₂2R₂

|i₁|R₁ = i₂R₂ → |i₁|i₂ = R₁R₂

i₂ = |i₁| · R₂R₁ = 0.75 · 0.100.15 = 0.50 A

B₂ = μ₀N₂i₂2R₂ = 4π·10⁻⁷ · 10 · 0.502 · 0.10 = 3.14·10⁻⁵ T

tan θ = B₂B_E∥ = 1.0 → B_E∥ = B₂ = 3.14·10⁻⁵ T

tan θ = μ₀N₁2R₁B_E∥ · i₁ + B₂B_E∥ שיפוע = 1.00.75 = 1.33 A⁻¹ → B_E∥ = 3.14·10⁻⁵ T

B_צפון = B_E∥ + B₂ = 2B_E∥ = 6.28·10⁻⁵ T tan θ = B₁B_E∥ + B₂ = B₁2B_E∥

tan θ = B₁2B_E∥ = 1.4 → B₁ = 2.8 · B_E∥ B₁ = 2.8 · 3.14·10⁻⁵ = 8.80·10⁻⁵ T

B₁ = μ₀N₁i₁2R₁ → i₁ = 2R₁B₁μ₀N₁ i₁ = 2 · 0.15 · 8.80·10⁻⁵4π·10⁻⁷ · 10 = 2.1 A

tan θ = 1.33 · i₁2 = 1.4 → i₁ = 2.81.33 = 2.1 A

E = |ΔV|Δd | V = E·d | C = QV = ε₀·Ad
½·m·v₀² = q·V | 1C_טור = 1C₁ + 1C₂ | C_מקביל = C₁ + C₂

15.0 mm

220

14667

5.91

1.30

1.46

88

2.95

0.43

1.30

E = |ΔV|Δd = 880.006 = 14667 V/m

V = E·d = 14667 · 0.015 = 220 V

C₁ = QV = 1.3·10⁻⁹220 = 5.9·10⁻¹² F = 5.9 pF

C = ε₀·Ad → A = C·dε₀ = 5.909·10⁻¹² · 0.0158.85·10⁻¹² = 1.0·10⁻² m²

F = q·E , q > 0 → F ‖ E

½·m·v₀² = q·V → v₀ = √2·q·Vm

v₀ = √2 · 3.2·10⁻¹⁹ · 2206.64·10⁻²⁷ = 1.5·10⁵ m/s

v₀ = √2·q·Vm

V = const , E = Vd ↑ , E·d = V → v₀ = const

E = Qε₀·A = const , V = E·d ↓ → v₀ ↓

C₂ = ε₀·½Ad = C₁2 = 2.95 pF

C_טור = C₁·C₂C₁ + C₂ = C₁·½C₁1.5·C₁ = C₁3 = 1.97 pF Q = C_טור·V = 1.97·10⁻¹² · 220 = 4.3·10⁻¹⁰ C = 0.43 nC

Q₁ = C₁·V = 5.909·10⁻¹² · 220 = 1.3·10⁻⁹ C = 1.3 nC

V₁ = V₂ = V = 220 V

Q₁Q₂ = C₁·VC₂·V = C₁C₂ = 2

ω = 314 s⁻¹ | I₀ = 2.42 A | r = 10 cm
N/L = 4000 m⁻¹ | R = 12 cm | המתח הרשום על הנורה 3 V

1

314

0.120

1

0.120

πr²

ε = −ΔΦΔt

Φ ∝ I₁ = I₀ sin(ωt)

ε ∝ −cos(ωt) → I₂ ∝ −cos(ωt)

B = μ₀NLI₁ = μ₀N I₀ sin(ωt)L

Φ = B · πr² = μ₀N I₀ πr² sin(ωt)L

ε₂ = −ΔΦΔt = −μ₀N I₀ πr² ωLcos(ωt) ε₂,max = μ₀NI₀πr²ωL

ε₂,max = 4π·10⁻⁷ · 4000 · 2.42 · π · (0.10)² · 314 = 0.12 V

ε_כולל = N_טבעת · ε₂,max

N_טבעת = 30.12 = 25

Φ = μ₀NLI₁ · πR² → ε₂,max = μ₀NLI₀πR²ω

ε₂,max = 4π·10⁻⁷ · 4000 · 2.42 · π · (0.12)² · 314 = 0.17 V

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