TS EAMCET · Physics · Capacitance
A fully charged capacitor has a capacitance \(C\). It is discharged through a small coil of resistance wire, embedded in a block of specific heat \(s\) and mass \(m\) under thermally isolated conditions. If the temperature of the block is raised by \(\Delta T\), the potential difference \(V\) across the capacitor initially is
- A \(\left(\frac{2 m s \Delta T}{C}\right)^2\)
- B \(\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}\)
- C \(\left(\frac{2 m s \Delta T}{C}\right)\)
- D \(2 m s \Delta T C\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Let \(V\) be the potential across the capacitor when it is fully charged, then, energy stored in the capacitor will be \(=\frac{1}{2} C V^2\) (where \(C=\) capacitance of capacitor) When the capacitor is fully discharged, loss of energy is in the form of heat \(=\Delta H\) As…
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