MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logic Gates

In the circui...

Question

In the circuit shown, key (K) is closed at $t = 0$ . Current through key at $t = 10^{- 3} l n (2)$ is

Open in App

Solution

$i_{1} = 2 (1 - e^{\frac{- R_{2} t}{L}}) \Rightarrow R_{2} = 10 Ω$

$q = C ε (1 - e^{\frac{- t}{R_{1} C}}) \Rightarrow R_{1} = 10 Ω$

$i_{2} = \frac{ε}{R_{1}} e^{\frac{- t}{R_{1} C}}$ $\frac{L}{R_{2}} = 10^{- 3}$

$i = i_{1} + i_{2} = 2 (1 - e^{- l n 2}) + 2 e^{- l n (2)}$

$i = 2 A$

Suggest Corrections

thumbs-up

0

Similar questions

Q. In the circuit shown below, the key

K

t = 0

. The current through the battery is

Q. In the circuit shown, the key (

K

t = 0

, the current through the key at the instant

t = 10^{- 3} ln (2)

Q. In the circuit shown, the key(K) is closed at t=0, the current through the key at the

t = 10^{- 1}

Q. In the circuit shown in the figure, key K is closed at

t = 0

. The current through the key at the instant

t = 10^{- 3} \times ln 2 s

Q. In the circuit show below,the key K is closed at t=0. the current through the battery is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Logic Gates

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon