A capacitor of capacitance is connected across an AC source of voltage , given by The displacement current between the plates of the capacitor, would then be given by [2021]

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Q.
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 [2021]

1 $I_{d} = V_{0} ω C \sin ω t$

2 $I_{d} = V_{0} ω C \cos ω t$

3 $I_{d} = \frac{V_{0}}{ω C} \cos ω t$

4 $I_{d} = \frac{V_{0}}{ω C} \sin ω t$

Ans.
(2)

The instantaneous voltage, $V = V_{0} \sin ω t$ ...(i)

Displacement current is given by

$i_{d} = \frac{C d V}{d t}$ ...(ii)

$i_{d} = \frac{C d}{d t} (V_{0} \sin ω t) \Rightarrow i_{d} = C V_{0} ω \cos ω t$

Solution

Q.
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 [2021]

1 $I_{d} = V_{0} ω C \sin ω t$

2 $I_{d} = V_{0} ω C \cos ω t$

3 $I_{d} = \frac{V_{0}}{ω C} \cos ω t$

4 $I_{d} = \frac{V_{0}}{ω C} \sin ω t$

Ans.
(2)

The instantaneous voltage, $V = V_{0} \sin ω t$ ...(i)

Displacement current is given by

$i_{d} = \frac{C d V}{d t}$ ...(ii)

$i_{d} = \frac{C d}{d t} (V_{0} \sin ω t) \Rightarrow i_{d} = C V_{0} ω \cos ω t$

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Solution

Q. 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 [2021]

1 Id=V0ωCsinωt

2 Id=V0ωCcosωt

3 Id=V0ωCcosωt

4 Id=V0ωCsinωt

Ans. (2) The instantaneous voltage, V=V0sinωt ...(i) Displacement current is given by id=CdVdt ...(ii) id=Cddt(V0sinωt) ⇒ id=CV0ωcosωt

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Q.
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 [2021]

1 $I_{d} = V_{0} ω C \sin ω t$

2 $I_{d} = V_{0} ω C \cos ω t$

3 $I_{d} = \frac{V_{0}}{ω C} \cos ω t$

4 $I_{d} = \frac{V_{0}}{ω C} \sin ω t$

Ans.
(2)

The instantaneous voltage, $V = V_{0} \sin ω t$ ...(i)

Displacement current is given by

$i_{d} = \frac{C d V}{d t}$ ...(ii)

$i_{d} = \frac{C d}{d t} (V_{0} \sin ω t) \Rightarrow i_{d} = C V_{0} ω \cos ω t$