From 4a14bbb9898977c7f634cf6b9dd44d6203f92cbf Mon Sep 17 00:00:00 2001 From: Renat Nurgaliyev Date: Mon, 22 Jun 2026 11:44:34 +0200 Subject: [PATCH] Drop answer-letter references from explanations --- explanations.json | 34 +++++++++++++++++----------------- 1 file changed, 17 insertions(+), 17 deletions(-) diff --git a/explanations.json b/explanations.json index 79e9f10..d558abb 100644 --- a/explanations.json +++ b/explanations.json @@ -5346,8 +5346,8 @@ "confidence": 8 }, "EA104": { - "revision": 3, - "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.", + "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); H/m is the unit of permeability instead.", "source": "https://50ohm.de/NEA_h_feld.html#EA104", "confidence": 8 }, @@ -5370,8 +5370,8 @@ "confidence": 8 }, "EA108": { - "revision": 4, - "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. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = $10^{-6}$.", + "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. The value $42 \\cdot 10^{-6}$ A is ten times too small. Hilfsmittel: use the Zehnerpotenzen/SI-Präfix table (S.11): micro = $10^{-6}$.", "source": "https://50ohm.de/NEA_zehnerpotenzen.html#EA108", "confidence": 8 }, @@ -5424,8 +5424,8 @@ "confidence": 8 }, "EA201": { - "revision": 3, - "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.", + "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). 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", "confidence": 8 }, @@ -5718,8 +5718,8 @@ "confidence": 8 }, "EB504": { - "revision": 4, - "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.) Hilfsmittel: combine $P = U\\cdot I$ with $I = U/R$ to get $P = U^2/R$, hence $U = \\sqrt{P\\cdot R}$ (Leistung, S.12).", + "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 $U = \\sqrt{P \\cdot R}$. The expression $\\sqrt{P/R}$ instead gives the current, not the voltage. Hilfsmittel: 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", "confidence": 8 }, @@ -6006,8 +6006,8 @@ "confidence": 8 }, "EC503": { - "revision": 3, - "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.", + "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. 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", "confidence": 8 }, @@ -6114,8 +6114,8 @@ "confidence": 8 }, "EC521": { - "revision": 4, - "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.) Hilfsmittel: apply $R = U/I$ (Ohmsches Gesetz, S.11) to the resistor's drop and Zener current.", + "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 $R = 8.8\\ \\text{V} / 0.030\\ \\text{A} \\approx 293\\ \\Omega$. A value of about $3.41\\,\\text{m}\\Omega$ would short the supply — a sanity-check fail. Hilfsmittel: 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", "confidence": 8 }, @@ -7224,8 +7224,8 @@ "confidence": 8 }, "EG305": { - "revision": 3, - "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.", + "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 absence of shielding does not itself prevent common-mode currents.", "source": "https://50ohm.de/NEA_uebertragungsleitungen_2.html#EG305", "confidence": 8 }, @@ -7350,8 +7350,8 @@ "confidence": 9 }, "EG502": { - "revision": 6, - "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 multiply 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). Hilfsmittel: 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).", + "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 multiply 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). Hilfsmittel: 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", "confidence": 8 }, @@ -10499,4 +10499,4 @@ "source": "https://50ohm.de/NEA_antennen_baurecht_haftung.html#VE707", "confidence": 8 } -} \ No newline at end of file +}