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Standard XII

Physics

Discharging of Capacitors

The current i...

Question

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?

40

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Solution

The correct option is D $48$ C

Given:
$I = 4 + 2 t$
Let dq be the charge which passes in a small interval of time dt. Then
$d q = I d t$
or $d q = (4 + 2 t) d t$
On integrating, we get
$q = \int_{2}^{6} (4 + 2 t) d t = [4 t + t^{2}]_{2}^{6} = 48$ C

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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 $t = 1 s$ .

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The current in a wire varies with time according to the equation I=4+2t, 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=2s to t=6s is?

The correct option is D 48CGiven:I=4+2tLet dq be the charge which passes in a small interval of time dt. Then dq=Idtor dq=(4+2t)dtOn integrating, we getq=∫62(4+2t)dt=[4t+t2]62=48C

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?

The correct option is D $48$ C

Given:
$I = 4 + 2 t$
Let dq be the charge which passes in a small interval of time dt. Then
$d q = I d t$
or $d q = (4 + 2 t) d t$
On integrating, we get
$q = \int_{2}^{6} (4 + 2 t) d t = [4 t + t^{2}]_{2}^{6} = 48$ C