text: # The capacitance Cx of a capacitive type sensor is (1000 x) pF, where x is the input to the sensor. As shown in the figure, the sensor is excited by a voltage 10 sin (100π t) V. The other terminal of the sensor is tied to the input of a high input impedance amplifier through a shielded cable, with shield connected to ground. The cable capacitance is 100 pF. The peak of the voltage VA at the input of the amplifier when x = 0.1 (in volts) is
  
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  ## Answer (Detailed Solution Below) 5
  
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  ## Detailed Solution
  
  Capacitance of a capacitive type sensor, Cx = (1000 x) pF
  
  When x = 0.1, Cx = (1000 × 0.1) pF = 100 pF
  
  Cable capacitance = 100 pF
  
  The figure shown below represents the electrical equivalent circuit of the given measurement system.
  
  As the amplifier has high input impedance there will not be any loading effect.
  
  Now, by applying the voltage division rule
  
  $${V_A} = {V_i} \times \frac{{\frac{1}{{j\omega {C_{x}}}}}}{{\frac{1}{{j\omega {C_{cable}}}} + \frac{1}{{j\omega {C_x}}}}}$$
  
  $${V_A} = 10 \times \frac{{{C_{cable}}}}{{{C_x} + {C_{cable}}}}$$
  
  $${V_A} = 10 \times \frac{{100 \times {{10}^{ - 12}}}}{{\left( {100 + 100} \right) \times {{10}^{ - 12}}}} = 5\;V$$