A capacitor C is charged by a battery of V volts. Then it is connected to an uncharged capacitor of capacity 2C as shown in figure. Now Match the following two columns.

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Solution

After closing switch both capacitors are in parallel so common potential is
$V_{c} = \frac{C V + 0}{C + 2 C} = \frac{V}{3}$
A) After closing the switch energy stored in C is $U_{C} = \frac{1}{2} C V_{c}^{2} = \frac{1}{2} C (V / 3)^{2} = \frac{1}{18} C V^{2}$ thus $A \to 3$
B) After closing the switch energy stored in 2C is $U_{2 C} = \frac{1}{2} (2 C) V_{c}^{2} = \frac{1}{2} (2 C) (V / 3)^{2} = \frac{1}{9} C V^{2}$ thus $B \to 1$
C) Loss of energy $= U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{2} (C + 2 C) V_{c}^{2} = \frac{1}{2} C V^{2} - \frac{1}{2} (3 C) (V / 3)^{2} = \frac{1}{2} C V^{2} - \frac{1}{6} C V^{2} = \frac{1}{3} C V^{2}$ thus $C \to 4$

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V

2 C

$(a)$ After closing the switch energy stored in $C$ .	$(p) \frac{1}{9} C V^{2}$
$(b)$ After closing the switch energy stored in $2 C$ .	$(q) \frac{1}{6} C V^{2}$
$(c)$ After closing the switch loss of energy during redistribution of charge.	$(r) \frac{1}{18} C V^{2}$
$(s)$ None of these