A capacitor of capacitance C- is connected across an AC source of voltage V- given by V - V0sin- t- The displacement current between the plates of the capacitor- would then be given by

Given V = V_0sinω t ...(1) Now, displacement current i_d is given by, I_d = C dVdt= C ddt (V_0 sinω t) nbsp; = C (V_0 ω) cosω t =V_0ω Ccosω t Hence ...

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Standard IX

Physics

Electric Field

A capacitor o...

Question

A capacitor of capacitance $C,$ is connected across an AC source of voltage $V,$ given by $V = V_{0} sin ω t$ . The displacement current between the plates of the capacitor, would then be given by

I_{d} = V_{0} ω C sin ω t

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I_{d} = V_{0} ω C cos ω t

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I_{d} = \frac{V_{0}}{ω C} cos ω t

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I_{d} = \frac{V_{0}}{ω C} sin ω t

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Solution

The correct option is B
$I_{d} = V_{0} ω C cos ω t$
Given $V = V_{0} sin ω t . . . (1)$

Now, displacement current $i_{d}$ is given by,

$I_{d} = C \frac{d V}{d t} = C \frac{d}{d t} (V_{0} sin ω t)$

$= C (V_{0} ω) cos ω t$

$= V_{0} ω C cos ω t$

Hence, $(B)$ is the correct answer.

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A capacitor of capacitance C, is connected across an AC source of voltage V, given by V=V0sinωt. The displacement current between the plates of the capacitor, would then be given by

The correct option is B Id=V0ωCcosω tGiven V=V0sinωt ...(1) Now, displacement current id is given by, Id=CdVdt=Cddt(V0sinωt) =C(V0ω)cosωt =V0ωCcosωt Hence, (B) is the correct answer.

A capacitor of capacitance $C,$ is connected across an AC source of voltage $V,$ given by $V = V_{0} sin ω t$ . The displacement current between the plates of the capacitor, would then be given by

The correct option is B
$I_{d} = V_{0} ω C cos ω t$
Given $V = V_{0} sin ω t . . . (1)$

Now, displacement current $i_{d}$ is given by,

$I_{d} = C \frac{d V}{d t} = C \frac{d}{d t} (V_{0} sin ω t)$

$= C (V_{0} ω) cos ω t$

$= V_{0} ω C cos ω t$

Hence, $(B)$ is the correct answer.