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Determine the potential difference between the points C and D in figure. |
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Answer» Solution :Since the capacitors are in series, charge on each capacitor is the same. At steady state, the charge on the capacitor is due to terminal patential differnee of cell `EPSILON`. When steady state is reached, no current FLOWS throught the capaciotrs. The current in the circuit is, `I = 12/( 1 + 3 +6)=1.2A` Terminal pot. diff. across A and B, `V = epsilon - I r= 12 -1.2 xx 1 = 10.8V` Capacitors are in series, their effective capacitance `C= ( 1 xx 2)/(1 +2) = 2/3 mu F` `:.` Charge on each capacitor, `q = 2/3 xx 10^(-6) xx 10.8 = 7.2 xx 10^(-6)C`. Pot. diff. between A and C `V_(A) - V_(C) = q/C_(2) = (7.2 xx 10^(-6))/(2 xx 10^(-6)) = 3.6 V` po. diff. between A and D `= V_(A) - V_(D) = IR_(2)` ` = 6 xx 1.2 = 7.2 V` Pot. diff. between C and D `= V_(C) - V_(D)` ` (V_(A) - V_(D)) - (V_(A) - V_(C))= 7.2 -3.6 = 3.6V` |
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