InterviewSolution
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InterviewSolution
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In the circuit shown, r = 4 Omega, C = 2mu F (a) Find the current coming out of the battery just after the switch is closed(b) Find charge on each capacitor in the steady state condition |
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Answer» Solution :(a) Due to symmetry. the points of equal potential are joined together, and the circuit may be reducedas `R_(13)=(r)/(2)+(r)/(4)+(r)/(4)+(r)/(2)=(3)/(2)r` `Here,r=4Omega,R_(13)=(3)/(2)(4)=6omega` THUS,`I=(24)/(R_(13)=(24)/(6)=4A` (b) in the steady-state condition, the circuit may be reduced as,` R_(13) = 2r = 2(4)Omega = 8Omega` `I=(24)/(8)=3A``V_(56)=(2r)(1)/(2)=Ir=(3)(4)=12V` Equivalent capacitance between 5 and 6 is `C_(56)=(C)/(2)=1muF` `thereforeq1_(56)=C_(56)V_(56)=12muC` NOW,`V_(97)=V_(17)-V_(19)` `=-r(1)/(2)+2(rl)/(2)=(rl)/(2)=((4)(3))/(2)=6V` therefore`q_(97)=(2)(6)=12muC` Similarly, `q_(89) = 12muC`  
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