text: # Solved Problems-11
  
  Problems-11
  A coil resonates at 2 MHz when a 18 pF capacitor is shunted across it. When shunting capacitor is 81 pF, the resonating frequency becomes 1 MHz. Find the distributed capacitor of the coil and the self resonating frequency.
  
  Solution
  Let the distributed capacitance of the coil be Cd (in pF).
  
  Show when the coil is shunted with another capacitance of C, the total capacitance becomes,
  (C + Cd).
  
  When C = 18 pF, the resonating frequency is f0 = 2 MHz
  (i)
  
  When C = 81 pF, the resonating frequency is f0 = 1 MHz
  
  (ii)
  
  Dividing (i) by (ii), we get,
  
  Putting the value of Cd in (ii), we get,
  
  So, the self-resonating frequency of the coil is given as,
explanation: The document contains a math problem and its solution. The problem involves finding the distributed capacitance of a coil and its self-resonating frequency based on given resonating frequencies and shunting capacitances. The solution is provided step-by-step, showing the calculations and the final answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Solved Problems-11
  
  Problems-11
  A coil resonates at 2 MHz when a 18 pF capacitor is shunted across it. When shunting capacitor is 81 pF, the resonating frequency becomes 1 MHz. Find the distributed capacitor of the coil and the self resonating frequency.
  
  Solution
  Let the distributed capacitance of the coil be Cd (in pF).
  
  Show when the coil is shunted with another capacitance of C, the total capacitance becomes,
  (C + Cd).
  
  When C = 18 pF, the resonating frequency is f0 = 2 MHz
  (i)
  
  When C = 81 pF, the resonating frequency is f0 = 1 MHz
  
  (ii)
  
  Dividing (i) by (ii), we get,
  
  Putting the value of Cd in (ii), we get,
  
  So, the self-resonating frequency of the coil is given as,
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem and its solution. The problem involves finding the distributed capacitance of a coil and its self-resonating frequency based on given resonating frequencies and shunting capacitances. The solution is provided step-by-step, showing the calculations and the final answer.
  
  Label: question-and-answer