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)
1.61 Wb (coming outside from the plane
similerly for side 2 (farther side of bigger circuit) applying same formula
b= 20cm = 0.02 m
C= 2.2m
a= 0.01 m
= 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 =
=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
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