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anonymous

  • one year ago

A capacitor is charged to a potential difference of 12V. it delivers 40% of its stored energy to a lamp. what is the final potential difference across the capacitor???

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  1. anonymous
    • one year ago
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    @Irishboy123

  2. anonymous
    • one year ago
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    @arindameducationusc

  3. arindameducationusc
    • one year ago
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    I think 12 is the charge....

  4. arindameducationusc
    • one year ago
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    U(Energy)=1/2*Q*V Q=charge V=potential difference (you have to find) U=40% of energy I think its this way but I am not sure, I am little weak in Electrostatics... but this should be the way....

  5. IrishBoy123
    • one year ago
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    \[E_1 = \frac{1}{2} C V_1^2 \ \] \[E_2 = \frac{1}{2}C V_2^2 = 0.60 E_1\] \[V_2^2 = 0.60 V_1 ^ 2\]

  6. anonymous
    • one year ago
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    How the formula E=1/2CV^2 CAME actually...i had able to derive E=CV^2 but from where did the 1/2 came???would be very kind of you to clear me..

  7. IrishBoy123
    • one year ago
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    look here first https://gyazo.com/ae43ccf0b318175dcb80738cbd06ba63 any questions, post them here! basic relationship is \[C = \frac{Q}{V}\] so however you derive the equation for energy, you can then manipulate it into other forms.....

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