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How to solve this problem- The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the potenti

The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second  the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be    

  • Option 1)

    200 and 180 V

  • Option 2)

    180 and 135 V

  • Option 3)

    160 and 80 V

  • Option 4)

    140 and 20 V

Answers (1)
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As we have learned

Potential Difference -

V=V_{0}\left ( 1-e^{\frac{-t}{Rc}} \right )

-

 

      V= V _0(e^{-\lambda t})
                     After 1 seconds

V_1= 320(e^(-\lambda )) \Rightarrow 240 = 320 (e^{-\lambda} ) \Rightarrow e^{-\lambda }= 3/4

V_2= 320(e^(-\lambda )^2) \Rightarrow 320 \times \left ( \frac{3}{4} \right ) ^2= 180 volt


 
 After 3 seconds V_3= 320(e^(-\lambda )^3) \Rightarrow 320 \times \left ( \frac{3}{4} \right ) ^3= 135 volt

 


Option 1)

200 and 180 V

Option 2)

180 and 135 V

Option 3)

160 and 80 V

Option 4)

140 and 20 V

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