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Question

Consider the situation shown in figure (31-E23). The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery. (c) Find the change in energy stored in the capacitors. (d) Find the heat developed in the system.

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Solution

Since the switch is opened for a long time, the capacitor is in series. ∴ ${C}_{eq}=\frac{C}{2}$ When the switch is closed, the charge flown from the battery is given by $q=\frac{\mathrm{C}}{2}×\epsilon =\frac{\mathrm{C}\epsilon }{2}$ (b) Work done, W$=q×V$ $⇒q=\frac{\mathrm{C}\epsilon }{2}×\epsilon =\frac{\mathrm{C}{\epsilon }^{2}}{2}$ (c) The change in the energy stored in the capacitors is given by $∆E=\frac{1}{2}{\mathrm{C}}_{eq}×{v}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×\frac{\mathrm{C}}{2}×{\epsilon }^{2}=\frac{1}{4}\mathrm{C}{\epsilon }^{2}$ (d) The heat developed in the system is given by $H=∆E\phantom{\rule{0ex}{0ex}}⇒H=\frac{1}{4}\mathrm{C}{\epsilon }^{2}$

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