ప్రతి సెకనుకు 200 జౌల్ వేడిని ఉత్పత్తి చేసినప్పుడు 2 ఓం నిరోధకంలో పొటెన్షియల్ బేధం _____ ద్వారా ఇవ్వబడుతుంది.

This question was previously asked in
RRB Group D 27 Sept 2022 Shift 3 Official Paper
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  1. 10 వోల్ట్
  2. 80 వోల్ట్
  3. 40 వోల్ట్
  4. 20 వోల్ట్

Answer (Detailed Solution Below)

Option 4 : 20 వోల్ట్
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Detailed Solution

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ఇచ్చినది:

వేడి ఉత్పత్తి =200 J.

నిరోధకం = 2 ఓం.

ఉపయోగించిన సూత్రం:

ఉత్పత్తి చేయబడిన వేడి మొత్తం \(H=\frac{V^2t}{R}\)

[ఇక్కడ V అనేది ఉత్పత్తి చేయబడిన వోల్టేజ్ మొత్తం; t అనేది అవసరమైన సమయం ;R అనేది నిరోధకం]

గణన:

    \(200=\frac{\mathrm{V}^2 \times 1}{2}\\ \Rightarrow \mathrm{V}^2=400 \\\Rightarrow \mathrm{V}=20 \mathrm{v}\)

కాబట్టి సరైన సమాధానం 20 వోల్ట్.

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