11. Advance Illustration | Current Electricity | Steady State Analysis of RC Circuits-3

11. Advance Illustration | Current Electricity | Steady State Analysis of RC Circuits-3


let us analyze another r c circuit in its
steady state. here we are, required to find the ratio of charges on 4 micro farad and
6 micro farad capacitor in steady state. here we know steady state no current flows through,
any capacitor. so here we just have a look if. there is any close loop where current
can flow. so here you can see in these 3 branches as capacitors are connected no current will
flow through these. so only in the outer loop current will circulate, and the value of this
current can be easily calculated, we can write current. in outer loop. is. here this current
can be given as total battery is. here, are of e m f 30 volt as both are, supporting each
other. so this is 30 divided by the total resistance of this loop as. 15 ohms. so this
will be 2 ampere so current 2 ampere is flowing through it. now to find out the charges on
these capacitor we take, at this point reference is zero volt. then, the potential on the other
side of this battery would be 18. so here, across this resistance the potential drop
will be i r, that is 2 multiplied by 5 10 ohms. so if current is flowing in this direction.
here potential will decrease. so 18 minus 10 this will be, 8 volt. here again 12 volt,
will be added so here potential will be 20 volt. and again, 2 ampere current passes thought
6 ohm resistance. so potential drop here will be 6 into 2 12. and 20 minus 12 here potential
is again, 8 volts. as we know the potential at all the points, here we can take the potential
at this junction as x. and we write nodal equation. at x. which can be written as sum
of the charges on these 3 plates of capacitor would be zero. as no potential drop is their
across this 3 ohm and 4 ohm resistance because no currents are flowing through these resistances.
here we can write charge on this plate will be 4 x. plus charge on this plate will be
2 multiplied by x minus, here potential is 8. this 2 multiplied by x minus 8 plus. here
potential is 6 multiplied by here potential will be x only, so this will also be x minus
8 is equal to zero. here, this gets cancelled out with 2. so we are getting 2 plus 1 3 plus
3 6 x is equal to. 8 multiplied by 3 24 plus 8 is 32. so the value of x will be 32 by 6
that is 16 by 3 volt. as we have got the value of x we can find out the charge on 4 micro
farad capacitor which can be given as 4 x. so this will be 4 multiplied by 16 by 3 this
is 64 by 3 micro coulomb. and charge on 6 micro farad capacitor we can write as 6 multiplied
by, x minus 8 as the value of x is less we can take it. other way that is 8 minus x which
i can write as 16 by 3. so on simplifying i am getting it. is equal to 2 multiplied
by, 8 that is 16 micro coulomb. so as, we asked in the question the ratio of, charge
on 4 micro farad to charge on, 6 micro farad capacitor we can write as 64 by 3. divided
by 16 so the result is 4 by 3. that is the, answer for this problem.

8 thoughts on “11. Advance Illustration | Current Electricity | Steady State Analysis of RC Circuits-3

  1. also sir these are all problems in dc power supply , how are problems in ac power supply solved,are they even in our iit course

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