A 10 mu F capacitor is charged by a 100 text{ | vedclass

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(D) Given: $C_1 = 10 \mu F$,$V_1 = 100 	ext{ V}$,$C_2 = 30 \mu F$,$V_2 = 0 	ext{ V}$.When two capacitors are connected,the electrostatic energy lost

Vedclass · Accepted Answer

$3.75 	imes 10^{-2} 	ext{ J}$

Vedclass · Answer

(D) Given: $C_1 = 10 \mu F$,$V_1 = 100 	ext{ V}$,$C_2 = 30 \mu F$,$V_2 = 0 	ext{ V}$.When two capacitors are connected,the electrostatic energy lost during the redistribution of charge is given by the formula:$\Delta U = \frac{1}{2} \left( \frac{C_1 C_2}{C_1 + C_2} ight) (V_1 - V_2)^2$Substituting the values:$\Delta U = \frac{1}{2} \left( \frac{10 	imes 10^{-6} 	imes 30 	imes 10^{-6}}{10 	imes 10^{-6} + 30 	imes 10^{-6}} ight) (100 - 0)^2$$\Delta U = \frac{1}{2} \left( \frac{300 	imes 10^{-12}}{40 	imes 10^{-6}} ight) (10000)$$\Delta U = \frac{1}{2} 	imes (7.5 	imes 10^{-6}) 	imes 10^4$$\Delta U = 3.75 	imes 10^{-2} 	ext{ J}$.

$A$ $10 \mu F$ capacitor is charged by a $100 \text{ V}$ battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu F$. During this process,the electrostatic energy lost by the first capacitor is

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