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Grade: 12
        
 
Some help on this question would be great
Thank you
7 months ago

Answers : (1)

Eshan
askIITians Faculty
1485 Points
							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}
2 months ago
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