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Drop answer-letter references from explanations

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Renat Nurgaliyev
2026-06-22 11:44:34 +02:00
parent 989d6d6888
commit 4a14bbb989
1 changed files with 17 additions and 17 deletions
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explanations.json
+16 -16
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@@ -5346,8 +5346,8 @@
"confidence": 8 "confidence": 8
}, },
"EA104": { "EA104": {
"revision": 3, "revision": 4,
"explanation": "Magnetic field strength $H$ is given in amperes per metre (A/m). The straight-wire case shows why: $H = I/(2\\pi r)$ is a current (amperes) spread over a circumference (metres). Keep $H$ (field strength, A/m) distinct from flux density $B$ (tesla); option D, H/m, is the unit of permeability instead.", "explanation": "Magnetic field strength $H$ is given in amperes per metre (A/m). The straight-wire case shows why: $H = I/(2\\pi r)$ is a current (amperes) spread over a circumference (metres). Keep $H$ (field strength, A/m) distinct from flux density $B$ (tesla); H/m is the unit of permeability instead.",
"source": "https://50ohm.de/NEA_h_feld.html#EA104", "source": "https://50ohm.de/NEA_h_feld.html#EA104",
"confidence": 8 "confidence": 8
}, },
@@ -5370,8 +5370,8 @@
"confidence": 8 "confidence": 8
}, },
"EA108": { "EA108": {
"revision": 4, "revision": 5,
"explanation": "Micro ($\\mu$) means $10^{-6}$. Expressing $0.00042$ A in microamperes shifts the decimal six places: $0.00042$ A $= 420 \\cdot 10^{-6}$ A $= 420\\ \\mu$A. Option D, $42 \\cdot 10^{-6}$ A, is ten times too small. <u>Hilfsmittel:</u> use the Zehnerpotenzen/SI-Präfix table (S.11): micro = $10^{-6}$.", "explanation": "Micro ($\\mu$) means $10^{-6}$. Expressing $0.00042$ A in microamperes shifts the decimal six places: $0.00042$ A $= 420 \\cdot 10^{-6}$ A $= 420\\ \\mu$A. The value $42 \\cdot 10^{-6}$ A is ten times too small. <u>Hilfsmittel:</u> use the Zehnerpotenzen/SI-Präfix table (S.11): micro = $10^{-6}$.",
"source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA108", "source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA108",
"confidence": 8 "confidence": 8
}, },
@@ -5424,8 +5424,8 @@
"confidence": 8 "confidence": 8
}, },
"EA201": { "EA201": {
"revision": 3, "revision": 4,
"explanation": "Binary needs only two symbols, 0 and 1, which map directly onto two robust electrical states — a switching element such as a transistor is simply off or on (low or high voltage). Multi-level analog states (option C) are far harder to keep reliable across temperature and noise, so digital logic builds everything from clean two-state switching.", "explanation": "Binary needs only two symbols, 0 and 1, which map directly onto two robust electrical states — a switching element such as a transistor is simply off or on (low or high voltage). Representing intermediate values between 0 and 1 with multi-level analog states is far harder to keep reliable across temperature and noise, so digital logic builds everything from clean two-state switching.",
"source": "https://50ohm.de/NEA_binaer.html#EA201", "source": "https://50ohm.de/NEA_binaer.html#EA201",
"confidence": 8 "confidence": 8
}, },
@@ -5718,8 +5718,8 @@
"confidence": 8 "confidence": 8
}, },
"EB504": { "EB504": {
"revision": 4, "revision": 5,
"explanation": "Start from $P = U \\cdot I$ and substitute Ohm's law $I = U/R$ to eliminate the unknown current: $P = U^2/R$. Solving for the voltage gives $U = \\sqrt{P \\cdot R}$. (Option C, $\\sqrt{P/R}$, actually gives the current, not the voltage.) <u>Hilfsmittel:</u> combine $P = U\\cdot I$ with $I = U/R$ to get $P = U^2/R$, hence $U = \\sqrt{P\\cdot R}$ (Leistung, S.12).", "explanation": "Start from $P = U \\cdot I$ and substitute Ohm's law $I = U/R$ to eliminate the unknown current: $P = U^2/R$. Solving for the voltage gives $U = \\sqrt{P \\cdot R}$. The expression $\\sqrt{P/R}$ instead gives the current, not the voltage. <u>Hilfsmittel:</u> combine $P = U\\cdot I$ with $I = U/R$ to get $P = U^2/R$, hence $U = \\sqrt{P\\cdot R}$ (Leistung, S.12).",
"source": "https://50ohm.de/NEA_leistung_2.html#EB504", "source": "https://50ohm.de/NEA_leistung_2.html#EB504",
"confidence": 8 "confidence": 8
}, },
@@ -6006,8 +6006,8 @@
"confidence": 8 "confidence": 8
}, },
"EC503": { "EC503": {
"revision": 3, "revision": 4,
"explanation": "Forward threshold (Schwellspannung) depends on the semiconductor: germanium conducts from about $0.2$-$0.4$ V, silicon from about $0.6$-$0.8$ V. The lower germanium drop is why crystal/detector receivers favour Ge or Schottky diodes for weak signals. Option B simply swaps the two materials.", "explanation": "Forward threshold (Schwellspannung) depends on the semiconductor: germanium conducts from about $0.2$-$0.4$ V, silicon from about $0.6$-$0.8$ V. The lower germanium drop is why crystal/detector receivers favour Ge or Schottky diodes for weak signals. Assigning $0.6$-$0.8$ V to germanium and $0.2$-$0.4$ V to silicon simply swaps the two materials.",
"source": "https://50ohm.de/NEA_diode_1.html#EC503", "source": "https://50ohm.de/NEA_diode_1.html#EC503",
"confidence": 8 "confidence": 8
}, },
@@ -6114,8 +6114,8 @@
"confidence": 8 "confidence": 8
}, },
"EC521": { "EC521": {
"revision": 4, "revision": 5,
"explanation": "With no load, the series resistor carries only the Zener current and must drop the supply down to the Zener voltage: $13.8\\ \\text{V} - 5\\ \\text{V} = 8.8$ V across it at $30$ mA. So $R = 8.8\\ \\text{V} / 0.030\\ \\text{A} \\approx 293\\ \\Omega$. (Option B's milliohm value would short the supply — a sanity-check fail.) <u>Hilfsmittel:</u> apply $R = U/I$ (Ohmsches Gesetz, S.11) to the resistor's drop and Zener current.", "explanation": "With no load, the series resistor carries only the Zener current and must drop the supply down to the Zener voltage: $13.8\\ \\text{V} - 5\\ \\text{V} = 8.8$ V across it at $30$ mA. So $R = 8.8\\ \\text{V} / 0.030\\ \\text{A} \\approx 293\\ \\Omega$. A value of about $3.41\\,\\text{m}\\Omega$ would short the supply — a sanity-check fail. <u>Hilfsmittel:</u> apply $R = U/I$ (Ohmsches Gesetz, S.11) to the resistor's drop and Zener current.",
"source": "https://50ohm.de/NEA_diode_1.html#EC521", "source": "https://50ohm.de/NEA_diode_1.html#EC521",
"confidence": 8 "confidence": 8
}, },
@@ -7224,8 +7224,8 @@
"confidence": 8 "confidence": 8
}, },
"EG305": { "EG305": {
"revision": 3, "revision": 4,
"explanation": "Open parallel-wire line (Hühnerleiter/Paralleldraht) runs mostly through air rather than a lossy solid dielectric, so it has markedly lower loss than coax and withstands high voltages well — ideal for feeding a high-SWR multiband antenna. Its drawback is that it radiates if unbalanced, the opposite of the shielding option B claims.", "explanation": "Open parallel-wire line (Hühnerleiter/Paralleldraht) runs mostly through air rather than a lossy solid dielectric, so it has markedly lower loss than coax and withstands high voltages well — ideal for feeding a high-SWR multiband antenna. Its drawback is that it radiates if unbalanced; the absence of shielding does not itself prevent common-mode currents.",
"source": "https://50ohm.de/NEA_uebertragungsleitungen_2.html#EG305", "source": "https://50ohm.de/NEA_uebertragungsleitungen_2.html#EG305",
"confidence": 8 "confidence": 8
}, },
@@ -7350,8 +7350,8 @@
"confidence": 9 "confidence": 9
}, },
"EG502": { "EG502": {
"revision": 6, "revision": 7,
"explanation": "Build EIRP in two steps. First get the real power at the antenna by subtracting feed-line losses from the transmitter output, $P_{\\text{Sender}} - P_{\\text{Verluste}}$; then <u>multiply</u> by the antenna gain factor $G$. Adding gain (options C/D) is the classic error — power and a linear gain factor multiply, not add (you would only add if everything were in dB). <u>Hilfsmittel:</u> subtract feed-line loss, then multiply by the gain factor: the sheet gives $P_\\mathrm{ERP} = P_S\\cdot 10^{(g_d-a)/(10\\,\\text{dB})}$ and $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1{,}64$ (S.15).", "explanation": "Build EIRP in two steps. First get the real power at the antenna by subtracting feed-line losses from the transmitter output, $P_{\\text{Sender}} - P_{\\text{Verluste}}$; then <u>multiply</u> by the antenna gain factor $G$. Adding antenna gain to power instead of multiplying by the linear gain factor is the classic error (you would only add if everything were in dB). <u>Hilfsmittel:</u> subtract feed-line loss, then multiply by the gain factor: the sheet gives $P_\\mathrm{ERP} = P_S\\cdot 10^{(g_d-a)/(10\\,\\text{dB})}$ and $P_\\mathrm{EIRP} = P_\\mathrm{ERP}\\cdot 1{,}64$ (S.15).",
"source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG502", "source": "https://50ohm.de/NEA_aequivalente_isotrope_strahlungsleistung_eirp_2.html#EG502",
"confidence": 8 "confidence": 8
}, },