A $600 p F$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 p F$ capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?

100, 6 \times 10^{- 6}

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200, 6 \times 10^{- 5}

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200, 5 \times 10^{- 6}

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100, 6 \times 10^{- 5}

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Solution

The correct option is B $100, 6 \times 10^{- 6}$
Total charge remains constant.
So, $q_{1} = q_{2}$
Common potential, $V_{C} = \frac{C_{1} V_{1}}{C_{1} + C_{2}} = 100 V$
Energy lost $= U_{i} - U_{f} = \frac{1}{2} C_{1} {V_{1}}^{2} - \frac{1}{2} (C_{1} + C_{2}) {V_{c}}^{2}$ $= 6 \times 10^{- 6} J$

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A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600pF capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?

The correct option is B 100,6×10−6Total charge remains constant. So, q1=q2Common potential, VC=C1V1C1+C2=100VEnergy lost =Ui−Uf=12C1V12−12(C1+C2)Vc2 =6×10−6J

A $600 p F$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $600 p F$ capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?

The correct option is B $100, 6 \times 10^{- 6}$
Total charge remains constant.
So, $q_{1} = q_{2}$
Common potential, $V_{C} = \frac{C_{1} V_{1}}{C_{1} + C_{2}} = 100 V$
Energy lost $= U_{i} - U_{f} = \frac{1}{2} C_{1} {V_{1}}^{2} - \frac{1}{2} (C_{1} + C_{2}) {V_{c}}^{2}$ $= 6 \times 10^{- 6} J$