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

Question

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. What is the common potential (in V) and energy lost (in J) after reconnection? (A) 100,6 × 10-6 (B) 200,6 × 10-5 (C) 200,5 × 10-6 (D) 100,6 × 10-5

Tardigrade Admin · Accepted Answer

VC=(C1 V1/C1+C2)=100 V Energy lost =Ui-Uf =(1/2) C1 V12-(1/2)(C1+C2) Vc2 =6 × 10-6 J

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

$100, 6 \times 1 0^{- 6}$

$200, 6 \times 1 0^{- 5}$

$200, 5 \times 1 0^{- 6}$

$100, 6 \times 1 0^{- 5}$

$V_{C} = \frac{C _{1} V _{1}}{C _{1} + C _{2}} = 100 V$
Energy lost $= U_{i} - U_{f}$
$= \frac{1}{2} C_{1} V_{1}^{2} - \frac{1}{2} (C_{1} + C_{2}) V_{c}^{2}$
$= 6 \times 1 0^{- 6} J$

Q. 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?

100,6×10−6

200,6×10−5

200,5×10−6

100,6×10−5