Two indentical capacitors each of capacitance 5 μ F are charged to potentials 2 kV and 1 kV respectively. The -ve ends are connected together. When the +ve ends are also connected together, the loss of energy of the system is

Question

Two indentical capacitors each of capacitance 5 μ F are charged to potentials 2 kV and 1 kV respectively. The -ve ends are connected together. When the +ve ends are also connected together, the loss of energy of the system is (A) 1.25 J (B) 5 J (C) 0 J (D) 160 J

Tardigrade Admin · Accepted Answer

Loss of energy = (1/2) (C1C2/(C1 + C2)) (V1 -V2)2 = (1/2) (5×10-6 ×5 × 10-6(2000-1000)2/(5+5)× 10-6) = (5× 5/2× 10) =1.25 J

Q. Two indentical capacitors each of capacitance $5 μ F$ are charged to potentials $2 kV$ and $1 kV$ respectively. The $- v e$ ends are connected together. When the $+ v e$ ends are also connected together, the loss of energy of the system is

1.25 J

5 J

0 J

160 J

Loss of energy $= \frac{1}{2} \frac{C _{1} C _{2}}{( C _{1} + C _{2} )} (V_{1} - V_{2})^{2}$
$= \frac{1}{2} \frac{5 \times 1 0 ^{- 6} \times 5 \times 1 0 ^{- 6} ( 2000 - 1000 ) ^{2}}{( 5 + 5 ) \times 1 0 ^{- 6}}$
$= \frac{5 \times 5}{2 \times 10} = 1.25 J$

Q. Two indentical capacitors each of capacitance 5μF are charged to potentials 2kV and 1kV respectively. The −ve ends are connected together. When the +ve ends are also connected together, the loss of energy of the system is

1.25 J

5 J

0 J

160 J

Loss of energy =21​(C1​+C2​)C1​C2​​(V1​−V2​)2 =21​(5+5)×10−65×10−6×5×10−6(2000−1000)2​ =2×105×5​=1.25J

Q. Two indentical capacitors each of capacitance $5 μ F$ are charged to potentials $2 kV$ and $1 kV$ respectively. The $- v e$ ends are connected together. When the $+ v e$ ends are also connected together, the loss of energy of the system is

Loss of energy $= \frac{1}{2} \frac{C _{1} C _{2}}{( C _{1} + C _{2} )} (V_{1} - V_{2})^{2}$
$= \frac{1}{2} \frac{5 \times 1 0 ^{- 6} \times 5 \times 1 0 ^{- 6} ( 2000 - 1000 ) ^{2}}{( 5 + 5 ) \times 1 0 ^{- 6}}$
$= \frac{5 \times 5}{2 \times 10} = 1.25 J$