KCET · Physics · Capacitance
The total energy stored in the condenser system shown in the figure will be
- A \(2 \mu \mathrm{J}\)
- B \(4 \mu \mathrm{J}\)
- C \(8 \mu \mathrm{J}\)
- D \(16 \mu \mathrm{J}\)
Answer & Solution
Correct Answer
(C) \(8 \mu \mathrm{J}\)
Step-by-step Solution
Detailed explanation
\(6 \mu \mathrm{F}\) and \(3 \mu \mathrm{F}\) capacitors are in series
\(\therefore \frac{1}{\mathrm{C}_{1}} =\frac{1}{6}+\frac{1}{3}\)
\(\mathrm{C}_{1} =2\)
\(\mathrm{C}_{1}\) is parallel to \(2 \mu \mathrm{F}\) capacitor
\(\therefore \mathrm{C}_{\mathrm{eq}}=2+2=4 \mu \mathrm{F}\)
Total energy, \(\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times(2)^{2}=8 \mu \mathrm{J}\)
\(\therefore \frac{1}{\mathrm{C}_{1}} =\frac{1}{6}+\frac{1}{3}\)
\(\mathrm{C}_{1} =2\)
\(\mathrm{C}_{1}\) is parallel to \(2 \mu \mathrm{F}\) capacitor
\(\therefore \mathrm{C}_{\mathrm{eq}}=2+2=4 \mu \mathrm{F}\)
Total energy, \(\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 4 \times(2)^{2}=8 \mu \mathrm{J}\)
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