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A capacitor of capacitance 1 microfarad is charged to 10 KV and then discharged through a wire. Find the heat produced in the wire in calorie.
1. 0.0119 calories
2. 1.19 calories
3. 11.9 calories
4. None of these

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Correct Answer - Option 3 : 11.9 calories

Concept of conversion of Joule to calories:

  • S.I unit of both Heat and Energy is the same i.e., Joule.
  • The CGS (Centimeter, Gram, Second) unit of Heat is calories.
  • 1 Calorie = 4.186 Joule.


We know that Energy stored by a capacitor is given by,

\({E_C} = \frac{1}{2}C{V^2}\) Joule

 \(H = \frac{{{E}}}{{4.186}}\) Calorie

Where H is Heat produced, E is Energy.

Calculation:

Given: C = 1 microfarad, VC = 10kv 

As per the formula,

\({E_C} = \frac{1}{2}C{V^2}\) Joule

\({E_C} = \frac{1}{2} \times \left( {1 \times {{10}^{ - 6}}} \right) \times {\left( {10 \times {{10}^3}} \right)^2}\)

EC = 50 Joule

\(H = \frac{{{E_C}}}{{4.186}}\)  Calorie

\(H = \frac{{50}}{{4.186}} = 11.9\;\) Calorie

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