RF ww 12.15 The circuit of Figure P12.15 will remove the DC portion of the input voltage, v (t), while amplifying the AC portion. Let vi (t) = 10 + 10-3 sin ot V, Rf = 10 k2, and Voatt = 20 V. a. Find...

For part(a) of the question, how is 10^-3sin(wt) got rid in the equation after "no DC in output..."

Also, why is the KCL eqn at the node : I_s+I_in
= I_finstead of I_s+I_f
= I_in? Does it matter whcih direction of current i choose?

Extracted text: RF ww 12.15 The circuit of Figure P12.15 will remove the DC portion of the input voltage, v (t), while amplifying the AC portion. Let vi (t) = 10 + 10-3 sin ot V, Rf = 10 k2, and Voatt = 20 V. a. Find Rs such that no DC voltage appears at the output. b. What is vout (1), using Rs from part a? Rs Voatt Vour(1) vị(1) Figure P12.15 (a:10K, b: 2×10* sinot) swers: (a) Inverting node: (V* - Vbatt)/Rs = (Vout-V)/RF V = (Vout Rs/RF +Vbatt) Rf/(Rs+RF) Non-inverting side: V* = V1 = 10+10³ sinøt Negative feedback V* = V (Vout Rs/RF +Vbatt) Rf/(Rs+Rf) = 10+10³sin@t Vout×Rs/RF = (10+10³ sinot) (Rs+Rf)/RF -Vbatt No DC in output, means: 10 (Rs+Rf)/RF -Vbatt Rs = RF (Vbatt/10 -1) = 10k (20/10-1) = 10K