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In one part of above network, steady current is flowing. Values of resistances are as shown in the diagram. Find energy stored in capacitor. |
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Answer» SOLUTION :When the capacitor becomes fully charged, there is no current in the branch CONTAINING capacitor. Now, current in the branch AD and DE is 1 + 2 = 3 A. Current in the branch EB is 1 A. Now, `V_(AD)= 3xx 5 = 15 ` V `V_(DE) = 3 xx 1 = 3 V ` ` V_(EB)= 1 xx 2= 2 `V Potential difference across capacitor is, `V = V_(AD) + V_(DE)+ V_(EB)` = 15 + 3 + 2 `therefore V = 20 ` Volt Now, electrostatic potential energy stored in the capacitor is, `U_(E)= (1)/(2) CV^(2)` `therefore U_(E) = (1)/(2) xx 4 xx 10^(-6) xx (20)^(2)` `thereforeU_(E)= 8 xx 10^(-4) `J |
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