The current in a wire varies with time according to the equation $I = 4 + 2 t$ , where $I$ is in $a m p e r e$ and $t$ is in $s e c$ . Calculate the quantity of charge which has passed through a cross-section of the wire during the time $t = 2 s e c$ to $t = 6 s e c$ .

Open in App

Solution

$I = \frac{d q}{d t} = 4 + 2 t$
$\int d q = \int (4 + 2 t) d t$
$q = 4 t + t^{2} = 6^{2} + 4 \times 6 - 2^{2} - 4 \times 2 = 48$

Suggest Corrections

Similar questions

Q. The current in a wire varies with time according to the equation

I = 4 + 2

t, where I is in ampere and t is in second. The quantity of charge which has to be passed through a cross-section of the wire during the time

t = 2

s to

t = 6

s is?

Q. The current in a conductor varies with time t as

I = 2 t + 3 t^{2}

amp. where I is ampere and t in second. Electric charge flowing through a section of the conductor during

t = 2 s e c

t = 3 s e c

is:-

Q. A current flowing through a wire depends on time as

I = 3 t^{2} + 2 t + 5

. The charge flowing through the cross section of the wire in time from

t = 0

t = 2 s e c

Q. A current in a wire is given by the equation,

I = 2 t^{2} - 3 t + 1

, the charge through cross section of wire in time interval

t = 3

s to

t = 5

s is?

The current through a wire depends on time as, $i = (10 + 4 t)$ Here, $i$ is in ampere and $t$ is in seconds. Find the charge that crosses through a section in time interval between $t = 0$ to $t = 1 s$ .

Join BYJU'S Learning Program

The current in a wire varies with time according to the equation I=4+2t, where I is in ampere and t is in sec. Calculate the quantity of charge which has passed through a cross-section of the wire during the time t=2 sec to t=6 sec.

I=dqdt=4+2t∫dq=∫(4+2t)dtq=4t+t2=62+4×6−22−4×2=48

The current in a wire varies with time according to the equation $I = 4 + 2 t$ , where $I$ is in $a m p e r e$ and $t$ is in $s e c$ . Calculate the quantity of charge which has passed through a cross-section of the wire during the time $t = 2 s e c$ to $t = 6 s e c$ .

$I = \frac{d q}{d t} = 4 + 2 t$
$\int d q = \int (4 + 2 t) d t$
$q = 4 t + t^{2} = 6^{2} + 4 \times 6 - 2^{2} - 4 \times 2 = 48$