An ideal coil of 10 H is joined in series with a resistance of 5 Ω and battery of 5 V, 2 s after joining the current flowing (in ampere) in the circuit will be

Question

An ideal coil of 10 H is joined in series with a resistance of 5 Ω and battery of 5 V, 2 s after joining the current flowing (in ampere) in the circuit will be (A)  e-1  (B)  (1-e-1)  (C)  (1-e)  (D) e

Tardigrade Admin · Accepted Answer

Given, L=10 H, R=5 Ω, E=5 V and t=2 s. Current in R-L circuit I=I0(1-e-R t / L) I=(E/R)(1-e-R t / L) =(5/5)(1-e-5 × 2 / 10) I=(1-e-1)

Q. An ideal coil of $10 H$ is joined in series with a resistance of $5 Ω$ and battery of $5 V, 2 s$ after joining the current flowing (in ampere) in the circuit will be

$e^{- 1}$

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

$(1 - e)$

$e$

Given, $L = 10 H, R = 5Ω, E = 5 V$
and $t = 2 s$ .
Current in $R - L$ circuit
$I = I_{0} (1 - e^{- Rt / L})$
$I = \frac{E}{R} (1 - e^{- Rt / L})$
$= \frac{5}{5} (1 - e^{- 5 \times 2/10})$
$I = (1 - e^{- 1})$

Q. An ideal coil of 10H is joined in series with a resistance of 5Ω and battery of 5V,2s after joining the current flowing (in ampere) in the circuit will be

e−1

(1−e−1)

(1−e)

e

Given, L=10H,R=5Ω,E=5V and t=2s. Current in R−L circuit I=I0​(1−e−Rt/L) I=RE​(1−e−Rt/L) =55​(1−e−5×2/10) I=(1−e−1)

Q. An ideal coil of $10 H$ is joined in series with a resistance of $5 Ω$ and battery of $5 V, 2 s$ after joining the current flowing (in ampere) in the circuit will be

$e^{- 1}$

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

$(1 - e)$

$e$

Given, $L = 10 H, R = 5Ω, E = 5 V$
and $t = 2 s$ .
Current in $R - L$ circuit
$I = I_{0} (1 - e^{- Rt / L})$
$I = \frac{E}{R} (1 - e^{- Rt / L})$
$= \frac{5}{5} (1 - e^{- 5 \times 2/10})$
$I = (1 - e^{- 1})$