Question

A capacitor of capacitt 5µF is charged up 220 volt and is disconnected from battery. Now charged 5µF capacitance is connected by another uncharged capacitor of 2.5µF capacitor . Find the heat loss in the process.

Answer 1

Answer:

• The Heat lost in the process is 10.09 × 10⁻² J

Given:

1. Charged Capacitor (C₁) = 5μF
2. Uncharged Capac. (C₂) = 2.5 μF

Explanation:

$\rule{300}{1.5}$

In Case-1 the capacitor 5μF is charged to a potential of 220 V.so, let's find the  energy stored in the capacitor 5μF.

$\large\bigstar\;\underline{\boxed{\sf U_{1}=\dfrac{1}{2}\; C_{1}V_{1}^{2}}}$

Here,

• U₁ Denotes energy.
• C₁ Denotes capacitance.
• V₁ Denotes voltage.

Substituting the values,

$\longmapsto\sf U_{1}=\dfrac{1}{2}\times 5\times 10^{-6}\times \bigg(220\bigg)^{2}\\\\\\\\\longmapsto\sf U_{1}=\dfrac{1}{2}\times 5\times 10^{-6}\times 48400\\\\\\\\\longmapsto\sf U_{1}=\dfrac{1}{2}\times 5\times 10^{-6}\times 10^{2}\times 484\\\\\\\\\longmapsto\sf U_{1}= 5\times 10^{-4}\times 242\\\\\\\\\longmapsto\sf U_{1}=1210\times 10^{-4}\\\\\\\\\longmapsto\sf U_{1}=12.10\times 10^{-2}\\\\\\\\\longmapsto\boxed{\sf U_{1}=12.10\times 10^{-2}\;J}\sf \quad\dots\dots(1)$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, when it is connected to uncharged the capacitor they both will reach a common potential.

And, we know that charge remains constant.

$\longmapsto\sf Q_{1}=Q_{2}\\\\\\\\\longmapsto\sf C_{1}\;.\;V_{1}=\bigg(C_{1}+C_{2}\bigg)\;.\;V_{2}$

• Here V₂ is the common potential.

$\longmapsto\sf 5\times 10^{-6}\times 220=\bigg(5+2.5\bigg)\times 10^{-6}\times V_{2}\\\\\\\\\longmapsto\sf 5\times 220=7.5\times V_{2}\\\\\\\\\longmapsto\sf V_{2}=\dfrac{5\times 220}{7.5}\\\\\\\\\longmapsto\sf V_{2}=\dfrac{2}{3}\times 220\\\\\\\\\longmapsto\sf V_{2}=\dfrac{440}{3} \\\\\\\\\longmapsto\underline{\underline{\sf V_{2}=\dfrac{440}{3}}}$

Now, lets find the energy stored in the combination of both capacitors.

$\large\bigstar\;\underline{\boxed{\sf U_{2}=\dfrac{1}{2}\; \bigg(C_{1}+C_{2}\bigg)\;.\;V_{2}^{2}}}$

Here,

• U₂ Denotes energy.
• C₁ & C₂ Denotes capacitance.
• V₂ Denotes voltage.

Substituting the values,

$\longmapsto\sf U_{2}=\dfrac{1}{2}\times \bigg(5+2.5\bigg)\times 10^{-6}\times \Bigg(\dfrac{220}{3}\Bigg)^{2}\\\\\\\\\longmapsto\sf U_{2}=\dfrac{1}{2}\times 7.5\times 10^{-6}\times \dfrac{48400}{9}\\\\\\\\\longmapsto\sf U_{2}=\dfrac{1}{2}\times 7.5\times 10^{-6}\times 10^{2}\times \dfrac{484}{9}\\\\\\\\\longmapsto\sf U_{2}= \dfrac{7.5}{9}\times 10^{-4}\times 242\\\\\\\\\longmapsto\sf U_{2}=\dfrac{1815}{9}\times 10^{-4}$

$\longmapsto\sf U_{2}=201.67\times 10^{-4}\\\\\\\\\longmapsto\sf U_{2}=2.01\times 10^{-2}\\\\\\\\\longmapsto\boxed{\sf U_{2}=2.01\times 10^{-2}\;J}\sf \quad\dots\dots(2)$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now let's find the heat lost in the process,

$\longmapsto\sf \Delta H=\Delta U=U_{1}-U_{2}\\\\\\\\\longmapsto\sf \Delta H=12.10\times 10^{-2}\;-\;2.01\times 10^{-2}\\\\\\\\\longmapsto\sf \Delta H=10.09\times 10^{-2}\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf \Delta H=10.09\times 10^{-2}\;J}}}}$

∴ The Heat lost in the process is 10.09 × 10⁻² J.

$\rule{300}{1.5}$

Answer 2

A capacitor of capacitance 5µF is charged up 220 volt and is disconnected from battery. Now charged 5µF capacitance is connected by another uncharged capacitor of 2.5µF capacitor.

To find : The heat loss in the process.

solution : first let's derive formula for this situation.

let C₁ is the capacitor charged up V volts and is disconnected from battery and connected to uncharged capacitor C₂

charge on 1st capacitor = C₁V

from conservation of charge,

C₁V = (C₁ + C₂)v'

⇒v' = C₁V/(C₁ + C₂)

now heat loss in the process = initial energy - final energy

= 1/2 C₁V² - 1/2 (C₁ + C₂)v'²

= 1/2 [C₁V² - (C₁ + C₂){C₁V/(C₁ + C₂)}² ]

= 1/2 C₁C₂V²/(C₁ + C₂), this is the required formula.

now putting, C₁ = 5 × 10¯⁶F , C₂ = 2.5 × 10¯⁶ F

and V = 220 V

so, ∆U = 1/2 (5 × 10¯⁶ )(2.5 × 10¯⁶) × (220)²/(7.5 × 10¯⁶)

= 1/2 × 5/3 × 48400 × 10¯⁶

= 121/3 × 10¯³ J

Therefore the heat loss in the process is 121/3 × 10¯³ J