An ideal coil of 10 H is connected in series with a resistance of 5 - and battery of 5 V- Two seconds after the connection is made the current flowing in amperes in the circuit is

The correct option is D (1 e^ 1) i_o=I_o(1 e^ 2× 510) =55(1 e^ 1) =(1 e^ 1) ...

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

Byju's Answer

Standard XII

Physics

Alternating Current Applied to an Inductor

An ideal coil...

Question

An ideal coil of $10$ $H$ is connected in series with a resistance of $5$ $Ω$ and battery of $5$ $V$ . Two seconds after the connection is made the current flowing in amperes in the circuit is

(1 - e)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

e^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(1 - e^{- 1})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $(1 - e^{- 1})$
$i_{o} = I_{o} (1 - e^{\frac{- 2 \times 5}{10}})$
$= \frac{5}{5} (1 - e^{- 1})$
$= (1 - e^{- 1})$

Suggest Corrections

Similar questions

Q. A coil of

10

H

inductance has a

5

Ω

resistance and is connected to a

5

V

battery in series. The current in ampere, in the circuit

2

seconds after the circuit is switched on is

Q. An ideal coil of

10 H

is connected in series with a resistance of

5 Ω

and a battery of

5 V .

2

second after the connection is made the current following in amperes in the circuit is:

Q. An ideal coil of

10

Henry is joined in series with a resistance of

5

ohm and a battery of

5

volt.

2

second after joining, the current flowing in ampere in the circuit will be :

Q. A coil of inductance

1 H

and resistance

100 Ω

is connected to a battery of emf

12 V .

Find the energy stored in the coil at an instant

t = 10 ms

after the circuit is switched on.
[Use

e^{- 1} = 0.367

]

Q. An inductor

L = 20 mH

, a resistor

R = 100 Ω

and a battery

E = 10 V

are connected in series. After a long time the battery is short-circuited. Find the current in the circuit,

1 ms

after short-circuiting.

[

Use

e^{- 5} = 0.00673]

Join BYJU'S Learning Program

An ideal coil of 10 H is connected in series with a resistance of 5 Ω and battery of 5 V. Two seconds after the connection is made the current flowing in amperes in the circuit is

The correct option is D (1−e−1)io=Io(1−e−2×510)=55(1−e−1)=(1−e−1)

An ideal coil of $10$ $H$ is connected in series with a resistance of $5$ $Ω$ and battery of $5$ $V$ . Two seconds after the connection is made the current flowing in amperes in the circuit is

The correct option is D $(1 - e^{- 1})$
$i_{o} = I_{o} (1 - e^{\frac{- 2 \times 5}{10}})$
$= \frac{5}{5} (1 - e^{- 1})$
$= (1 - e^{- 1})$