Question 122: AC fundamentals

Question

AC voltage V(t) = 20 sin$$\omega$$t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m
2
. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $$\varepsilon $$
0
= 8.85 $$	imes$$ 10
$$-$$12
F/m]

Accepted Answer

Given,
AC voltage, V(t) = 20 sin $$\omega$$t volt.
Frequency, f = 50Hz
Separation between the plates, d = 2 mm = 2 $$	imes$$ 10
$$-$$3
m
Area, A = 1 m
2
As, $$C = {{{\varepsilon _0}A} \over d}$$
where, $${{\varepsilon _0}}$$ = absolute electrical permittivity of free space = 8.854 $$	imes$$ 10
$$-$$12
N
$$-$$1
kg
2
m
$$-$$2
$$C = {{{\varepsilon _0} 	imes 1} \over {2 	imes {{10}^{ - 3}}}}$$ .... (i)
Capacitive reactance $$({X_C}) = {1 \over {\omega C}}$$ .... (ii)
From Eqs. (i) and (ii), we get
$${X_C} = {{2 	imes {{10}^{ - 3}}} \over {2 	imes 50\pi  	imes {\varepsilon _0}}}$$ ($$\because$$ $$\omega$$ = 2$$\pi$$f)
$$ = {{2 	imes {{10}^{ - 3}}} \over {25 	imes 4\pi {\varepsilon _0}}}$$
$$ \Rightarrow {X_C} = {{2 	imes {{10}^{ - 3}}} \over {25}} 	imes 9 	imes {10^9}$$
$$ \Rightarrow {X_C} = {{18} \over {25}} 	imes {10^6}\,\Omega $$
By using Ohm's law,
As, $${I_0} = {{{V_0}} \over {{X_C}}} = {{20 	imes 25} \over {18}} 	imes {10^{ - 6}} = 27.78 	imes {10^{ - 6}}$$
$$\Rightarrow$$ I
0
= 27.78$$\mu$$A
$$	herefore$$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $$\mu$$A.

Answer

55.58 $$\mu$$A

Answer

21.14 $$\mu$$A

Answer

27.79 $$\mu$$A

Answer

83.37 $$\mu$$A

AC fundamentals

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