An ideal coil of 10 text H is connected in series with a resistance of 5 Ω and a battery of 5 text V . 2 seconds after the connection is made, the current flowing in ampere in the circuit is

Question

An ideal coil of 10 text H is connected in series with a resistance of 5 Ω and a battery of 5 text V . 2 seconds after the connection is made, the current flowing in ampere in the circuit is (A) (1-e-1) (B) (1 - textite) (C)  textite (D)  textite-1

Tardigrade Admin · Accepted Answer

During the growth of current in L R circuit is given by I=I0(1-e-(R/L) t) or I=(E/R)(1-e-(R/L) t)=(5/5)(1-e-(5/10) × 2) I=(1-e-1)

Q. An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . $2$ seconds after the connection is made, the current flowing in ampere in the circuit is

$(1 - e^{- 1})$

$(1 - e)$

$e$

$e^{- 1}$

During the growth of current in $L R$ circuit is given by
$I = I_{0} (1 - e^{- \frac{R}{L} t})$
or $I = \frac{E}{R} (1 - e^{- \frac{R}{L} t}) = \frac{5}{5} (1 - e^{- \frac{5}{10} \times 2})$
$I = (1 - e^{- 1})$

Q. An ideal coil of 10 H is connected in series with a resistance of 5Ω and a battery of 5 V . 2 seconds after the connection is made, the current flowing in ampere in the circuit is

(1−e−1)

(1−e)

e

e−1

During the growth of current in LR circuit is given by I=I0​(1−e−LR​t) or I=RE​(1−e−LR​t)=55​(1−e−105​×2) I=(1−e−1)

Q. An ideal coil of $10 H$ is connected in series with a resistance of $5 Ω$ and a battery of $5 V$ . $2$ seconds after the connection is made, the current flowing in ampere in the circuit is

$(1 - e^{- 1})$

$(1 - e)$

$e$

$e^{- 1}$

During the growth of current in $L R$ circuit is given by
$I = I_{0} (1 - e^{- \frac{R}{L} t})$
or $I = \frac{E}{R} (1 - e^{- \frac{R}{L} t}) = \frac{5}{5} (1 - e^{- \frac{5}{10} \times 2})$
$I = (1 - e^{- 1})$