IIT JEE , AIEEE Forum - Physics , Magnetism

trisha

A circuit is as described below.

consider three vertical lines.In the 1st line a battery of emf E(with +ve plate upwards & -ve plate downwards) is connected.In the 2nd line a capacitor A & an inductor L are connected such that the inductor is below the capacitor.In the 3rd line another capacitor B & a switch S₂ are connected.now the top most points of all the 3 lines are joined by a horizontal line.while joining the bottom most points of these 3 lines a switch S₁ is placed between the 1st & 2nd lines.

Now at t=0,the switch S₁ is closed while switch S₂ remains open.At t=t_o=(LC)^1/2pi/2, switch S₂is closed while switch S₁is opened.

Now 1)what will be the charge on capacitor A after time t_o?

2)what is the current flowing through the inductor at t=t_o?

3)After switch S₂ is closed and S₁is opened,what will be the maximum value of current through the inductor?

rajat

1. charge = CE

2. zero

3. 2(pie)Esqroot(2C/L)

yahiyafirdous

Let current flowing through the circuit at any time is i

i=C dv_c/dt

E=Ldi/dt +(1/C)i.dt

Ld²i/dt² + i/C=0

d[(di/dt)²]/dt= -(i/LC)di/dt

(di/dt)²= -i²/LC + K where K is a constant

At t=0,v_L(0)=Ldi(0)/dt = E, i(0)=0

K=(E/L)²

(di/dt)²=(E/L)² - i²/LC

Solving above diff. equation:

i=E(C/L) sin[t/(LC) + K'/(LC)]

At t=0, i(0)=0

Hence K'=0

i=E(C/L) sin[t/(LC)]

(1)

At t=t₀=(/2)(LC)

i=E(C/L)_{[ 0]}^{[ t₀]}sin[t/(LC)] dt= EC Ans

(2)

At t=t₀=(/2)(LC)

i=E(C/L)sin[t/(LC)] = E(C/L) Ans

(3)

When S₁ is open and S₂ closed, only frequency will change , but max. current will be same as in previous case.

i_max=E(C/L) Ans.

trisha

I understood 1,2 answers given by you.But answer for 3rd question given is ^{[ ]}

3/^{[ ]}

2 times the answer which u got.so please verify it

rajat

trisha .what r the answers ??

for the first question yahiyafirdous' answer is a sine term !

ok,mathematically u may get this solution but i just dont see how and why the current will oscilate .after all the capacitor is charging so why will the current oscillate

yahiyafirdous

Yes you are correct, the max. current in the circuit after S₂ closed is

E(3C/2L), proceed as follows:

d²i/dt²+ (2/LC)i=0

Solving for di/dt we get:[ at t=0, v_L(0)= -E, i(0)=E(C/L), get integration constant=3(E/L)²]

di/dt=[3(E/L)² - 2i²/LC]

For i to be maximum, di/dt=0

i_max=E(3C/2L), Ans

NOTE: ASK IF ANY DOUBT

^{[ ]}

trisha

I cant understand how u got the integration constants ( i mean how u got V_L=E=Ldi/dt, i.e., di/dt=E/L)in the first and third case also.please explain me in detail yahiafiradous.

please help me yahiafiradous, i will definetely vote u and give u a salute.

Please also tell me how u got d²i/dt²+2i/LC=0 in the third question

trisha

Please also tell me how u got d²i/dt²+2i/LC=0 in the third case

rajat

but what r the answers ??

yahiyafirdous

S₂ closed and S₁ open, so circuit con tains two C and one L

v_L=Ldi/dt v_L=voltage across L, v_c= voltage across C
i=C dv_c/dt
v_c=1/C

idt
Apply Kirchoff's loop law:

v_L+2v_c=0
Differentiate with respect to t

dv_L/dt + 2 dv_c/dt =0
Now put v_L=Ldi/dt , i=C dv_c/dt in above equation, you will get
d²i/dt²+2i/LC=0
Now it is obvious, only mathematical solution is required.

NOTE: I am not here to collect rating score, just I am here to help u people.

trisha

But my doubt is how u got V_L=E in the third question?

Please explain me why it is

krishna.gopal

Let me solve this question from begining.

By KVL

E=L(di/dt)+(1/C)_{[ 0]}

^{[t ]} tdt

diff once again

(d²i/dt²)=-i/(LC)

Sol of this problem is of the form

i=Asin(wt)+Bcos(wt) where w^2 = 1/(LC)

at t=0, i=0, therefore B=0

At t=0 di/dt = E/L

Aw=E/L So A = Esqrt(C/L)

a) At t= t_o=pi/(2w)

Q =₀

^toAsin(wt)dt =-(A/w)[cos(wt_o)-cos(0)]

cos(wt_o) = cos(pi/2)=0

Q=A/w =CE

b) i(t₀) =Asin(pi/2) = A = Esqrt(C/L)

c) After t=t₀lets define a new time such that closing of switch 2 is t=0

differential equation will become

(d²i/dt²)=-2i/(LC)

which will have a solution of form

i=Csin(w't)+Dcos(w't) where w'^2=2/(LC)

At t=0 i=Esqrt(C/L)

From last part it can be seen that di/dt=0 at this point

S0 C=0 and D= Esqrt(C/L)

So maximum current will remain same as earlier

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