Question 5: AC fundamentals

Question

An ideal coil of $$10 \mathrm{~H}$$ is connected in series with a resistance of $$5 \Omega$$ and a battery of $$5 \mathrm{~V}$$. After $$2 \mathrm{~s}$$, after the connection is made, the current flowing (in ampere) in the circuit is

Accepted Answer

Rise of current in $$L-R$$ circuit is given by
$$I=I_0\left(1-e^{-t / 	au}ight)$$
where, $$\quad I_0=\frac{E}{R}=\frac{5}{5}=1 \mathrm{~A}$$
Now, $$	au=\frac{L}{R}=\frac{10}{5}=2 \mathrm{~s}$$
After $$2 \mathrm{~s}$$, i.e. at $$t=2 \mathrm{~s}$$
Rise of current, $$I=\left(1-e^{-1}ight) \mathrm{A}$$

Answer

$$(1-e)$$

Answer

$$e$$

Answer

$$e^{-1}$$

Answer

$$\left(1-e^{-1}\right)$$

AC fundamentals

An ideal coil of $10 \mathrm{~H} is connected in series with a resistance of 5 \Omega and a battery of 5 \mathrm{~V}. After 2 \mathrm{~s}$, after the connection is made, the current flowing (in ampere) in the circuit is

Choose the correct answer: