KCET · Physics · Capacitance
Two identical capacitors each of capacitance \(5 \mu \mathrm{F}\) are charged to potentials \(2 \mathrm{kV}\) and \(1 \mathrm{kV}\) respectively, Their -ve ends are connected together. When the +ve ends are also connected together, the loss of energy of the system is
- A 160 J
- B zero
- C 5 J
- D 1.25 J
Answer & Solution
Correct Answer
(D) 1.25 J
Step-by-step Solution
Detailed explanation
\(\text {Loss of energy } =\frac{1}{2} \frac{C_{1} C_{2}}{\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2} \)
\( = \frac{1}{2} \frac{5 \times 10^{-6} \times 5 \times 10^{-6}(2000-1000)^{2}}{(5+5) \times 10^{-6}} \)
\( = \frac{5 \times 5}{2 \times 10}=1.25 \mathrm{~J}\)
\( = \frac{1}{2} \frac{5 \times 10^{-6} \times 5 \times 10^{-6}(2000-1000)^{2}}{(5+5) \times 10^{-6}} \)
\( = \frac{5 \times 5}{2 \times 10}=1.25 \mathrm{~J}\)
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