For the circuit given below, the value of the g11 and g21 are _________________(a) g11 = –\(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(b) g11 = \(\frac{1}{R_1-R_2}\), g21 = –\(\frac{R_2}{R_1+R_2}\)(c) g11 = \(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(d) g11 = \(\frac{1}{R_1-R_2}\), g21 = \(\frac{R_2}{R_1-R_2}\)I got this question in examination.Origin of the question is Advanced Problems Involving Parameters in chapter Two-Port Networks of Network Theory

For the circuit given below, the value of the g11 and g21 are ________

1.	For the circuit given below, the value of the g11 and g21 are _________________(a) g11 = –\(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(b) g11 = \(\frac{1}{R_1-R_2}\), g21 = –\(\frac{R_2}{R_1+R_2}\)(c) g11 = \(\frac{1}{R_1+R_2}\), g21 = \(\frac{R_2}{R_1+R_2}\)(d) g11 = \(\frac{1}{R_1-R_2}\), g21 = \(\frac{R_2}{R_1-R_2}\)I got this question in examination.Origin of the question is Advanced Problems Involving Parameters in chapter Two-Port Networks of Network Theory
Answer» The CORRECT option is (C) G11 = \(\frac{1}{R_1+R_2}\), G21 = \(\frac{R_2}{R_1+R_2}\) To explain I would SAY: I1 = \(\frac{V_1}{R_1+R_2}\) Or, g11 = \(\frac{I_1}{V_1}= \frac{1}{R_1+R_2}\) By voltage division, V2 = \(\frac{R_2}{R_1+R_2}\)V1 Or, g21 = \(\frac{V_2}{V_1} = \frac{R_2}{R_1+R_2}\).

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