2.27 A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

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alok kumar singh | Contributor-Level 10

3 months ago

2.27 Capacitance of the charged capacitor, $C_{1}$ = 4 $μ F = 4 \times 10^{- 6}$ F

Supply voltage, $V_{1}$ = 200 V

Electrostatic energy stored in the $C_{1}$ capacitor, $E_{1}$ = $\frac{1}{2} C_{1} {V_{1}}^{2}$ = $\frac{1}{2} \times 4 \times 10^{- 6} \times 200^{2}$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_{2}$ = 2 $μ F$ = 2 $\times 10^{- 6}$ F

When $C_{2}$ is connected to $C_{1}$ , the potential acquired by $C_{2}$ be $V_{2}$

From the conservation of energy, the charge acquired by $C_{1}$ becomes the charge acquired by $C_{1}$ and $C_{2} .$

Hence $V_{2} \times (C_{1} + C_{2}) = C_{1} V_{1}$

or $V_{2}$ = $\frac{C_{1} V_{1}}{(C_{1} + C_{2})}$ = $\frac{4 \times 10^{- 6} \times 200}{(4 \times 10^{- 6} + 2 \times 10^{- 6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_{2}$ =

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2.27 Capacitance of the charged capacitor, $C_{1}$ = 4 $μ F = 4 \times 10^{- 6}$ F

Supply voltage, $V_{1}$ = 200 V

Electrostatic energy stored in the $C_{1}$ capacitor, $E_{1}$ = $\frac{1}{2} C_{1} {V_{1}}^{2}$ = $\frac{1}{2} \times 4 \times 10^{- 6} \times 200^{2}$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_{2}$ = 2 $μ F$ = 2 $\times 10^{- 6}$ F

When $C_{2}$ is connected to $C_{1}$ , the potential acquired by $C_{2}$ be $V_{2}$

From the conservation of energy, the charge acquired by $C_{1}$ becomes the charge acquired by $C_{1}$ and $C_{2} .$

Hence $V_{2} \times (C_{1} + C_{2}) = C_{1} V_{1}$

or $V_{2}$ = $\frac{C_{1} V_{1}}{(C_{1} + C_{2})}$ = $\frac{4 \times 10^{- 6} \times 200}{(4 \times 10^{- 6} + 2 \times 10^{- 6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_{2}$ = $\frac{1}{2}$ ( $C_{1} + C_{2}) \times V_{2}^{2}$ = $\frac{1}{2} \times$ ( $4 \times 10^{- 6} +$ 2 $\times 10^{- 6}) \times {(\frac{400}{3})}^{2}$ = 5.33 $\times 10^{- 2}$ J

Hence the amount of electrostatic energy lost by capacitor $C_{1} = E_{1} - E_{2}$

= 0.08 - 5.33 $\times 10^{- 2}$ J = 0.0267 J = 2.67 $\times 10^{- 2}$ J

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2.27 A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

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