Two resistors of 10 ohm and 20 ohm and an ideal inductor of 10H are connected to a 2v battery. The key is shorted at time t=0. Find the initial and final currents through the battery.

 Hi,

 The resistances of 10 and 20 ohm are connected in series hence the equivalent resistance is hence 30 ohm = R

Inductance is given as 10 ohm …..(given)

Now potential difference across R = IR

Potential difference across L = LdI/dt

We have to find the minimum and maximum current. When current reaches its maximum value, dI/dt will be = 0.

Hence let the value of maximum current be I₀.

Therefore, at I = I₀ , dI/dt = 0,

Emf of the battery (E) will be:

E = IR +LdI/dt

For I = I₀

E = I₀R

E = 2v as given in the problem

Hence substituting in the above equation:

2 = I₀R

2 = I₀30

Therefore I₀ = 2/30 = 1/15A this is the final current to be calculated.

From the growth of current in inductive circuit eqn , we know that: I = I₀(1-e^-(R/L)t)

In the problem the initial time is given as t = 0 applying tis in the above eqn we get,

I = I₀(1-e⁰)

I = 0 which is the initial current which needs to be calculated.