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The resistance each of 16 Omega and capacitance of each 100 mu F are arranged as shown in the figure. A battery of emf 12 V is joined across A and B. Find the (i) reading of the ammeter just after key is ciosed and after long time. (ii) charges in each capacitors when steady state is achieved. |
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Answer» Solution :(`i`) Just after the key is closed The circuit will be as shown in figure `1` Then circuit can be simplify as figure `2`  The points `D` and `E` will be at same potential hence the circuit can be further simplified as shown in figure `3` & `4`  Hence equivalent RESISTANCE across `A` and `B`, `R_(eq)=((68)/(7)xx16)/((68)/(7)+16)=6.04 Omega` Hence reading of ammeter `I=(12)/(6.4)~~2A`. After long time the circuit can be redrawn as  `R_(AB)=(16xx32)/(16+32)=(32)/(3)Omega` Hence the reading of the ammeter `I=(9)/(8)A` (`ii`) The current through paths `ADC` and `AEC` will be same and will be equal to `I_(2)=(1)/(2)xX(9)/(8)xx(16)/(48)=(3)/(16)A` Hence potential difference across `AD`, `AE`, `DC` and `CE`, `V'=16xx(3)/(16)=3V` Hence the charge in the `C_(1)`, `C_(2)`, `C_(5)` and `C_(6)q=100xx3=300muC` ALSO `V_(DB)=V_(EB)=12-3=9V` Hence the charge in the capacitor `C_(3)` and `C_(4)=100xx9=900muC` |
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