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Grade 12General Physics

Some help on this question would be great
Thank you

Question image for Some help on this question would be great Thank y
Profile image of qwerty
8 Years agoGrade 12
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1 Answer

Profile image of Eshan
8 Years ago
Decrease in potential energy of the capacitor=\dfrac{1}{2}C(V^2-(\dfrac{V}{2})^2)=\dfrac{3CV^2}{8}

Work done on the battery= Voltage of battery times the charge that went into it.

=\dfrac{V}{2}\times C(V-\dfrac{V}{2})=\dfrac{CV^2}{4}
Hence amount of heat dissipated=\dfrac{3CV^2}{8}-\dfrac{CV^2}{4}=\dfrac{CV^2}{8}