TS EAMCET · Physics · Capacitance
Two capacitors of capacities \(1 \mu \mathrm{F}\) and \(C \mu \mathrm{F}\) are connected in series and the combination is charged to a potential difference of \(120 \mathrm{~V}\). If the charge on the combination is \(80 \mu \mathrm{C}\), the energy stored in the capacitor of capacity \(C\) in \(\mu \mathrm{J}\) is
- A 1800
- B 1600
- C 14400
- D 7200
Answer & Solution
Correct Answer
(B) 1600
Step-by-step Solution
Detailed explanation
Capacitances \(1 \mu \mathrm{F}\) and \(C \mu \mathrm{F}\) are connected in series, then \(C_{\text {eq }}=\frac{C}{1+C}\) Given, \(V=120 \mathrm{~V}\) and \(q=80 \mu \mathrm{C}\) \(\because \quad q=C_{\mathrm{eq}} V\) \(80=\frac{C}{C+1} \times 120\) or…
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