a battery of 10 volt is connected to a capacitor of capacitance 0.1 if the battery is now removed and this capacitor is connected

Question

a battery of 10 volt is connected to a capacitor of capacitance 0.1 if the battery is now removed and this capacitor is connected to a parallel with another uncharged capacitor of capacitance 0.2 how much of electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation​

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Anna 2 years 2021-06-22T03:16:34+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-06-22T03:18:14+00:00

    Answer:

    Explanation:

    Capacitance of a charged capacitor, C  

    1

    ​  

    =4μF=4×10  

    −6

    F

    Supply voltage, V  

    1

    ​  

    =200V

    Electrostatic energy stored in C1 is given by,

    E  

    1

    ​  

    =  

    2

    1

    ​  

    C  

    1

    ​  

    V  

    1

    2

    ​  

     

         =  

    2

    1

    ​  

    ×4×10  

    −6

    ×(200)  

    2

     

         =8×10  

    −2

    J

    Capacitance of an uncharged capacitor, C  

    2

    ​  

    =2μF=2×10  

    −6

    F  

    When C  

    2

    ​  

     is connected to the circuit, the potential acquired by it is V  

    2

    ​  

    .

    According to the conservation of charge, initial charge on capacitor C  

    1

    ​  

     is equal to the final charge on capacitors, C  

    1

    ​  

     and C  

    2

    ​  

     

    ∴V  

    2

    ​  

    (C  

    1

    ​  

    +C  

    2

    ​  

    )=C  

    1

    ​  

    V  

    1

    ​  

     

    V  

    2

    ​  

    ×(4+2)×10  

    −6

    =4×10  

    −6

    ×200

    V  

    2

    ​  

    =  

    3

    400

    ​  

    V

    Electrostatic energy for the combination of two capacitors is given by,

    E  

    2

    ​  

    =  

    2

    1

    ​  

    (C  

    1

    ​  

    +C  

    2

    ​  

    )V  

    2

    2

    ​  

     

        =  

    2

    1

    ​  

    (2+4)×10  

    −6

    ×(  

    3

    400

    ​  

    )  

    2

     

        =5.33×10  

    −2

    J

    Hence, amount of electrostatic energy lost by capacitor C  

    1

    ​  

     

    =E  

    1

    ​  

    −E  

    2

    ​  

     

    =0.08−0.0533=0.0267

    =2.67×10  

    −2

    J

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