2. Electric Potential and Capacitance
hard

Two Identical capacttors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ of equal capacitance are connected as shown in the circult. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $\mathrm{C}_{1}$ using battery of $emf \;V\; volt$. Now disconnecting $a$ and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?.....$\%$

A

$75$

B

$0$

C

$50$

D

$25$

(NEET-2019)

Solution

$U_{\text {initial }}=\frac{1}{2} C V^{2},$ Loss $=\frac{C \cdot C}{2(C+C)}(V-0)^{2}=\frac{1}{4} C V^{2}$

$\%$ Loss $=\frac{\frac{1}{4} \mathrm{CV}^{2}}{\frac{1}{2} \mathrm{CV}^{2}} \times 100=50 \%$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.