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???

Question

anonymous · Answer

@Irishboy123

anonymous · Answer

@arindameducationusc

arindameducationusc · Answer

I think 12 is the charge....

arindameducationusc · Answer

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....

irishboy123 · Answer

$$E_1 = \frac{1}{2} C V_1^2 \ $$

$$E_2 = \frac{1}{2}C V_2^2 = 0.60 E_1$$

$$V_2^2 = 0.60 V_1 ^ 2$$

anonymous · Answer

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..

irishboy123 · Answer

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.....