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```        In the circuit given in the figure the initial value of current through the battery on closing the circuit(i.e. K is pressed) is
ans: 0.3 A
```
7 years ago

28 Points
```										Dear Siddharth,
Ans:- I think your ans is wrong why Let me explain
Let the current in the inductor is i1 and through the parallel resistor is i2
Then from kirchoff,s laws we get
10(i1+i2)+10(i1+i2)+10 i2=6..............(1)
L di1/dt =10i2
or 5 di1/dt=10i2...............(2)
Solving these two eq we get
3/5=2i1+3/2di1/dt
integrating from i1 varying from 0 to i1 we get the current through the inductance is
i1=0.3(1-exp-4t/3)
hence initially at t=0, i1=0 i.e no current flows through the inductance
So, from eq (1) we get i2=0.2 amp
Hence the current through the battery is = 0.2 amp (ans)

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```
7 years ago
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