Question

In the circuit shown in the figure below, the capacitance has a value of C = 20uF and is initially charged to 100 V with the polarity shown. The resistor Ro = 101. At t=0, the switch is closed. The small circuit is not connected to the large circuit in any way. The wire of the small circuit has a resistance of 1.0/m and contains 25 turns. The large circuit is a rectangle 2.0 m by 4.0 m while the small circuit has dimensions a = 10.0 cm and b=20.0 cm. The distance cis 5.0 cm (the figure is very clearly not drawn to scale...). Both circuits are held stationary, assume that only the wire nearest the small coil produces an appreciable magnetic field through the coil. (a) [1 point] What is the current in the large circuit 200 us after the switch Sis closed? (b) [1 point] What is the current in the small circuit 200 us after the switch Sis closed? (c) [1 point] What is the direction of the current in the small circuit?

Answer 1

given data

C = 20 micro F

initial voltage= 100 V

maximim charge on cacacitor Q= CV= 0.002 C

equation of voltge decay, current genration and charge decay when capacitor is discharging-(derivation is at last if required)

V = Voe-(t/RC)    and also I = Ioe-(t/RC)    Q = Qoe-(t/RC)

I = Ioe-(t/RC) Io= Vo/R

I = 10 e-(0.0002/10X0.00002)

I = 10/e

I = 3.678 A..................................ans 1

PART 2

the total resistance of circuit 2 is

total length is = 2 (a + b) X number of turn

                       = 2X30 X 25

                       = 1500 cm = 15m

total resistance = 15 X 1 = 15 ohm

the genration of current in second circuit is because of induction. as the current is changing in circuit 1 so the magnetic field around it will change.so there will be induction in second circuit

the flux change in the smaller circuit due to 1 (branch of bigger circuit as shown in figure below)

”

0, = 5 Bjadr = leta Hoilo dr

$\frac{4XpiX10^{-7}X3.678X20X10^{-2}}{2pi} ln(\frac{5+10}{5})$

1.61 Wb (coming outside from the plane

similerly for side 2 (farther side of bigger circuit) applying same formula

0, = 5 Bjadr = leta Hoilo dr

b= 20cm = 0.02 m

C= 2.2m

a= 0.01 m

$\frac{4XpiX10^{-7}X3.678X20X10^{-2}}{2pi} ln(\frac{2.2+0.01}{2.2})$

= 0.00667 Wb (inside to the plane of smaller circuit)

TOtal change in flux will be = 1.61-0.0067

                                           = 1.603 Wb

induced voltage = $\frac{d\phi }{dt}$

                            =1.603/0.0002

                               8016.5 V

current = V/R

              = 8016.5/10

                          = 801.6 Amp

     

the direction of current in the circuit is anti clock wise

because the current is decresing ..so do the magnetic flux link with smaller circuit decreases (comming outward the plane)and the induced current flow like will it ncreases current

The current (I) in the discharge at that instant is therefore:
I = - dq/dt

But V = IR and q = CV so dq/dt = d(CV)/dt = C dV/dt
Therefore we have V = -CR dV/dt Rearranging and integrating gives:

Sorry for calculation mistakes if any