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In the circuit shown, r=4Omega, C=2muF (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 equalpotential are joined together, and the circuit MAY be reduced as `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)(I)/(2)=Ir=(3)(4)=12V` Equivalent CAPACITANCE between `5` and `6` is `C_(56)=(C )/(2)=1muF` `:. q_(56)=C_(56)V_(56)=12muC` Now, `V_(97)=V_(17)-V_(19)= -r(I)/(2)+2(rI)/(2)=(rI)/(2)=((4)(3))/(2)=6V` `:. q_(97)=CV_(97)=(2)(6)=12muC` Similarly, `q_(89)=12muC` 
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