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In, all the capacitors are in steady state initially. i. What is the charge flowing through the switch when it is closed? ii. What is the charge flowing section AB? iii. What is the work done by the battery? iv. What is the heat produced when (S) is closed? . |
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Answer» Solution :Before closing the SWITCH, `4x + (x-6) 2+ (x-6)2=0` or `x=4 V` `|Q_(4muF)|=16 muC,|Q_(2muF)|=4muC` `|Q_(6muF)|=12 muC` When switch `S_(w)` is closed, the `4 muF` capacuit is short-circuited, and the potential difference across it BECOMES zero. The circuit after closing the switch is shown in   .  . shown the charges coming and going out of the capacitors. So the charge through section `AB` is `32 muC`, and the charge through the switch is `48 muC`. When the switch is closed, `16 muC` charge from the left plate goes the switch is charge PASSING through th battery is `32 muC`. so the work done by the battery is `W_("battery")=DeltaqV=(32 muC)(6V)=192 mu J` Heat produced is `H=W_("battery")-DeltaU=W_("battery")-(U_(f)U_(i))` `U_(f)=1/2(2)xx6^(2)+1/2(6)(6)^(2)=144 mu J` and `U_(i)=1/(2)4(4)^(2)+1/2(8)2^(2)=48 muJ` `H=[192-(144-48)]mu J=96 mu J`. |
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