When a switch is closed- the current built up in an electric circuit is given by E- Ri - Ldidt- If L - 640-R-250-E-500 and i-0 at t-0 -then to what does the current approach as t -

The correct option is D 2 ampGiven Differential equation is E=Ri+L di / dt ,Separating the Variables and integrating gives, ∫ _ 0 ^ i 1 E iR ...

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

Byju's Answer

Standard XII

Physics

Kirchhoff's Junction Law

When a switch...

Question

When a switch is closed, the current built up in an electric circuit is given by $E = R i + L \frac{d i}{d t}$ . If $L = 640, R = 250, E = 500$ and $i = 0$ at $t = 0$ ,then to what does the current approach as $t \to \infty$

2.5

amp

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

2

amp

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

3

amp

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

4

amp

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

Open in App

Solution

The correct option is D $2$ amp
Given Differential equation is $E = R i + L \frac{d i}{d t}$ ,
Separating the Variables and integrating gives,
$\int_{0}^{i} \frac{1}{E - i R} d i = \int_{0}^{t} \frac{1}{L} d t$ ,

$\frac{1}{R} ln (\frac{E}{E - i R}) = t / L$ ,
$i = \frac{E}{R} (1 - e^{- R t / L})$ ,
As $t ⟶ \infty ⟹ e^{- R t} ⟶ 0$ ,
$i = t ⟶ \infty ⟹ e^{- R t} ⟶ 0$ ,
$∴ i = \frac{500}{250} = 2 a m p$ .

Suggest Corrections

Similar questions

Q. In the circuit shown in the following figure

E = 10 V, R_{1} = 2 ohm, R_{2} = 3 ohm

and

R_{3} = 6 ohm

and

L = 5 henry

. The current

I_{1}

just after pressing the switch

S

(correct answer + 2, wrong answer - 0.50)

Q. State whether the following statement is True or False.
The current in the circuit with inductance

L

and resistance

R

and voltage

E s i n ω t

is given by

L \frac{d i}{d t} + R i = E s i n ω t

.if

i = 0

t = 0

then,current is given by

i = \frac{E}{\sqrt{E^{2} + L^{2} ω^{2}}} [s i n (ω t - ϕ) + e^{- R t / L} s i n ϕ]

where

ϕ = t a n^{- 1} (\frac{L ω}{R})

Q. Consider the circuit shown in the figure. The current i₃ is equal to

Q. The current in the given circuit is increasing with a rate

a = 4

amp/s. The charge on the capacitor at an instant when the current in the circuit is

2

amp will be:

The conducting spheres of radius r and 2r at very large separation. Each sphere is having charge Q. These spheres are connected with a conducting wire of resistance R. Then, the current flowing in the wire at $t = 2 μ s$ . (Where r = 30cm, $R = 90 k Ω$ and $Q = 10 μ C)$ )

Join BYJU'S Learning Program

When a switch is closed, the current built up in an electric circuit is given by E=Ri+Ldidt. If L=640,R=250,E=500 and i=0 at t=0 ,then to what does the current approach as t→∞

The correct option is D 2 ampGiven Differential equation is E=Ri+Ldidt,Separating the Variables and integrating gives,∫i01E−iRdi=∫t01Ldt,1Rln(EE−iR)=t/L,i=ER(1−e−Rt/L),As t⟶∞⟹e−Rt⟶0,i=t⟶∞⟹e−Rt⟶0,∴i=500250=2amp.

When a switch is closed, the current built up in an electric circuit is given by $E = R i + L \frac{d i}{d t}$ . If $L = 640, R = 250, E = 500$ and $i = 0$ at $t = 0$ ,then to what does the current approach as $t \to \infty$