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.
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)
66
1@kaymant sir: yes,that`s how i have done....though i dnt know if my calculations are correct or not....
1
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