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Loss Of Energy In Redistribution Of Charges | Redistribution Of Charges

Loss Of Energy In Redistribution Of Charges | Redistribution Of Charges :- When capacitors charged to different potentials are connected with each other, charge flows from the capacitor at higher potential to the one at lower potential. In this process, although the total charge is conserved, some energy is lost in the form of heat due to the flow of charge through the connecting wires. To determine this energy loss, consider two capacitors having capacitances C₁ and C₂, which are charged to potentials V₁ and V₂ respectively, and are then connected in parallel. The positive plates of both capacitors are connected together, and the negative plates are also connected together.

After charge transfer, the common potential (V) ,

$\displaystyle V=\frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ …..(1)

Total energy before combination,

$\displaystyle {{U}_{1}}=\frac{\text{1}}{2}{{C}_{1}}V_{1}^{2}+\frac{\text{1}}{2}{{C}_{2}}V_{2}^{2}$ …..(2)

Total energy after the combination,

$\displaystyle {{U}_{2}}=\frac{\text{1}}{2}{{C}_{1}}{{V}^{2}}+\frac{\text{1}}{2}{{C}_{2}}{{V}^{2}}=\frac{\text{1}}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{V}^{2}}$ …..(3)

Using equation (1) in equation (3), we get

$\displaystyle {{U}_{2}}=\frac{\text{1}}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{\left( \frac{{{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)}^{2}}$

$\displaystyle \Rightarrow {{U}_{2}}=\frac{\text{1}}{2}\frac{{{\left( {{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} \right)}^{2}}}{{{C}_{1}}+{{C}_{2}}}$ …..(4)

Let’s use equations (2) and (4) to find the value of (U₁ – U₂) :-

$\displaystyle {{U}_{1}}-{{U}_{2}}=\frac{\text{1}}{2}{{C}_{1}}V_{1}^{2}+\frac{\text{1}}{2}{{C}_{2}}V_{2}^{2}-\frac{\text{1}}{2}\frac{{{\left( {{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} \right)}^{2}}}{{{C}_{1}}+{{C}_{2}}}$

$\displaystyle {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}V_{1}^{2}\left( {{C}_{1}}+{{C}_{2}} \right)+{{C}_{2}}V_{2}^{2}\left( {{C}_{1}}+{{C}_{2}} \right)-{{\left( {{C}_{1}}{{V}_{1}}+{{C}_{2}}{{V}_{2}} \right)}^{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle {{U}_{1}}-{{U}_{2}}=\frac{C_{1}^{2}V_{1}^{2}+{{C}_{1}}{{C}_{2}}V_{1}^{2}+{{C}_{1}}{{C}_{2}}V_{2}^{2}+C_{2}^{2}V_{2}^{2}-C_{1}^{2}V_{1}^{2}-C_{2}^{2}V_{2}^{2}-2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \Rightarrow {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}{{C}_{2}}\left( V_{1}^{2}+V_{2}^{2}-2{{V}_{1}}{{V}_{2}} \right)}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \therefore {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}{{C}_{2}}{{\left( {{V}_{1}}-{{V}_{2}} \right)}^{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$ …..(5)

In the above equation (5), the value of (U₁ – U₂) is positive, i.e., (U₁ > U₂).

This proves that when charged capacitors are connected with each other, there is a loss of energy in redistribution of charges. This energy loss occurs in the connecting wires in the form of heat and sparking.

Note :-

If the positive plate of the first capacitor is connected to the negative plate of the second capacitor, and the negative plate of the first capacitor is connected to the positive plate of the second capacitor, then the energy loss :

After charge transfer, the common potential (V) is :

$\displaystyle V=\frac{{{C}_{1}}{{V}_{1}}-{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}}$ …..(6)

Total energy before combination,

$\displaystyle {{U}_{1}}=\frac{\text{1}}{2}{{C}_{1}}V_{1}^{2}+\frac{\text{1}}{2}{{C}_{2}}V_{2}^{2}$ …..(7)

Total energy after the combination,

$\displaystyle {{U}_{2}}=\frac{\text{1}}{2}{{C}_{1}}{{V}^{2}}+\frac{\text{1}}{2}{{C}_{2}}{{V}^{2}}=\frac{\text{1}}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{V}^{2}}$ …..(8)

Using equation (6) in equation (8), we get

$\displaystyle {{U}_{2}}=\frac{\text{1}}{2}\left( {{C}_{1}}+{{C}_{2}} \right){{\left( \frac{{{C}_{1}}{{V}_{1}}-{{C}_{2}}{{V}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)}^{\text{2}}}=\frac{\text{1}}{2}\frac{\left( C_{1}^{2}V_{1}^{2}+C_{2}^{2}V_{2}^{2}-2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}} \right)}{\left( {{C}_{1}}+{{C}_{2}} \right)}$ …..(9)

Using equations (7) and (9), the value of the energy loss (U₁ – U₂) is :-

$\displaystyle {{U}_{1}}-{{U}_{2}}=\frac{\text{1}}{2}{{C}_{1}}V_{1}^{2}+\frac{\text{1}}{2}{{C}_{2}}V_{2}^{2}-\frac{\text{1}}{2}\frac{\left( C_{1}^{2}V_{1}^{2}+C_{2}^{2}V_{2}^{2}-2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}} \right)}{\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \Rightarrow {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}V_{1}^{2}\left( {{C}_{1}}+{{C}_{2}} \right)+{{C}_{2}}V_{2}^{2}\left( {{C}_{1}}+{{C}_{2}} \right)-\left( C_{1}^{2}V_{1}^{2}+C_{2}^{2}V_{2}^{2}-2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}} \right)}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \Rightarrow {{U}_{1}}-{{U}_{2}}=\frac{C_{1}^{2}V_{1}^{2}+{{C}_{1}}{{C}_{2}}V_{1}^{2}+{{C}_{1}}{{C}_{2}}V_{2}^{2}+C_{2}^{2}V_{2}^{2}-C_{1}^{2}V_{1}^{2}-C_{2}^{2}V_{2}^{2}+2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \Rightarrow {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}{{C}_{2}}V_{1}^{2}+{{C}_{1}}{{C}_{2}}V_{2}^{2}+2{{C}_{1}}{{C}_{2}}{{V}_{1}}{{V}_{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \Rightarrow {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}{{C}_{2}}\left( V_{1}^{2}+V_{2}^{2}+2{{V}_{1}}V \right)}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

$\displaystyle \therefore {{U}_{1}}-{{U}_{2}}=\frac{{{C}_{1}}{{C}_{2}}{{\left( {{V}_{1}}+{{V}_{2}} \right)}^{2}}}{2\left( {{C}_{1}}+{{C}_{2}} \right)}$

Loss Of Energy In Redistribution Of Charges | Redistribution Of Charges

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