The input to the circuit shown below is νi = 2 sin ωt

Given that, V_i = 2 sin ωt mV

V₀ = –	10	. V_i → (1)
	1

KCL at node B

	V₀	+	V₀ – V_A	= i₀
	4		10

V_A = 0
(There4; due to virtual ground)

or	V₀	+	V₀	= i₀
	4		10

or	–10V_i	+	10V_i	= i₀
	4		10

or i₀ = – 3.5 V_i
or i₀ = – 3.5.2 sin ωt µA
or i₀ = – 7 sin ωt µA