1@kaymant sir: yes,that`s how i have done....though i dnt know if my calculations are correct or not....
Find the charge on 2μF capacitor in steady state :
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7 Answers
262Q= 20 uC
(potential drop across left 4uF capacitor =0. hence VD across 2uF=10 V. then Q=CV)
66Hint: In steady state the capacitors won't allow the current. In the present case, there is no current in any of the resistances and you can just remove them.
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i think its 60uC
A,B AT SAME POTENTIAL B,D AT SAME POTENTIAL B,C AT SAME POTENTIAL
SO A,C & C,D AT SAME POTENTIAL
SO THE REMAINING CIRCUIT IS 4uC &2uC IN PARALLEL
WITH PD 10V
1@chinton:yup ur method is right....i`ve made some pretty terrible silly mistakes....
71To Everyone.. Yes I have tried that all but they are giving the answer as 10μC