An ideal choke of 10 H is joined in series with resistance of 5 - and a battery of 5 V- The current in the circuit in 2 s after joining in ampere will be -

The correct option is D 1 e ^ 1 Time constant = L R = 10 5 =2 s ie, is 2 s , the current sizes to ( 1 1 e #160;) or ( 1 e ...

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Standard XII

Physics

Ohm's Law in Scalar Form

An ideal chok...

Question

An ideal choke of $10 H$ is joined in series with resistance of $5 Ω$ and a battery of $5 V$ . The current in the circuit in $2 s$ after joining in ampere will be :

e^{- 1}

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1 - e^{- 1}

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1 - e

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Solution

The correct option is D $1 - e^{- 1}$
Time constant $= \frac{L}{R} = \frac{10}{5} = 2 s$
ie, is $2 s$ , the current sizes to $(1 - \frac{1}{e})$
or $(1 - e^{- 1})$ times.

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Question 26
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 $Ω$ when connected to a 10 V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10 $Ω$ is connected in parallel with this series combination, what change (if any) in current flowing through 5 $Ω$ conductors and the potential difference across the lamp will take place? Give reason.

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An ideal choke of 10H is joined in series with resistance of 5Ω and a battery of 5V. The current in the circuit in 2s after joining in ampere will be :

The correct option is D 1−e−1Time constant =LR=105=2sie, is 2s, the current sizes to (1−1e)or (1−e−1) times.

An ideal choke of $10 H$ is joined in series with resistance of $5 Ω$ and a battery of $5 V$ . The current in the circuit in $2 s$ after joining in ampere will be :

The correct option is D $1 - e^{- 1}$
Time constant $= \frac{L}{R} = \frac{10}{5} = 2 s$
ie, is $2 s$ , the current sizes to $(1 - \frac{1}{e})$
or $(1 - e^{- 1})$ times.