The following circuit has R = 10k&ohm:, C = 10&m;F. The

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The following circuit has R = 10k&ohm:, C = 10&m;F. The input voltage is a sinusoid at 50Hz with an rms value of 10V under ideal conditions, the current i_s from the source is

1. 10 π mA leading by 90°
2. 20 π mA leading by 90°
3. 10 mA leading by 90°
4. 10 π mA lagging by 90°

Correct Option: D

X_c =	1	=	1	=	10³	ohm
	ω C		2π × 50 × 10μF		π

V_A =	- jX_c	. V₀
	-jX_c + R

∴ V₀ =	- jX_c + R	. V_A
	-jX_c

Assuming virtual ground,
V_in = V_A = 10V

∴ V₀ =	- jX_c + R	. 10
	-jX_c

Current , i_s =	V_in - V₀
	R

=	1		10 -	-jX_c + R	.10
	10⁴			-jX_c

=	1		1 -	-jX_c + R
	10³			-jX_c

=	1			-jX_c + jX_c - R
	10³			-jX_c

=	1		R	=	1	.	1	×	10 × 10³
	10³		jX_c		j		10³		(10³ / π)

=	10π × 10^-3	= (10π mA) ∠ -90° lagging
	j

Alternately
V₁ = V_s = R . i_s + V₀

V₂ = V_o		1
		jωC		= V_o	1

		1			+ R	1 + jωRC
		jωC

But V_o = A(V₁ – V₂) = A		V_s -	V₀
			1 + jωRC

⇒ V_o		1 +	A		= AV_s
			1 + jωRC

⇒	V₀	=	A(1 + jωRC)	≃ 1 + jωRC
	V_s		1 + jωRC + A

= 1 + j 100 π × 10⁴ × 10 × 10^–6 = 1 + j 10π
⇒ V_o = V_s (1 + j 10π) = 10 (1 + j 10π)
But V_s = R . i_s + V_o

or i_s =	V_s - V₀	=	10 - 10 - j100π
	R		10 × 10³

= – j 10π mA = 10π ∠– 90° mA