Physics-
General
Easy

Question

In the circuit shown in the figure, the ac source gives a voltage V equals 20 cos invisible function application left parenthesis 2000 rightwards arrow over short leftwards arrow t right parenthesis. Neglecting source resistance, the voltmeter and ammeter reading will be

  1. 0V, 0.47A    
  2. 1.68V, 0.47A    
  3. 0V, 1.4 A    
  4. 5.6V, 1.4 A    

The correct answer is: 5.6V, 1.4 A


    Z equals square root of left parenthesis R right parenthesis to the power of 2 end exponent plus left parenthesis X subscript L end subscript minus X subscript C end subscript right parenthesis to the power of 2 end exponent semicolon end root
    R equals 10 capital omega comma X subscript L end subscript equals omega L equals 2000 cross times 5 cross times 1 0 to the power of negative 3 end exponent equals 10 capital omega
    X subscript C end subscript equals fraction numerator 1 over denominator omega C end fraction equals fraction numerator 1 over denominator 2000 cross times 50 cross times 1 0 to the power of negative 6 end exponent end fraction equals 10 capital omega i. e. Z equals 10 capital omega
    Maximum current i subscript 0 end subscript equals fraction numerator V subscript 0 end subscript over denominator Z end fraction equals fraction numerator 20 over denominator 10 end fraction equals 2 A
    Hence i subscript r m s end subscript equals fraction numerator 2 over denominator square root of 2 end fraction equals 1.4 A
    and V subscript r m s end subscript equals 4 cross times 1.41 equals5.64 V

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