Question

For the circuit given below, the value of the g12 and g22 are _______________

(a) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 + \(\frac{R_1 R_2}{R_1+R_2}\)

(b) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 + \(\frac{R_1 R_2}{R_1+R_2}\)

(c) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 – \(\frac{R_1 R_2}{R_1+R_2}\)

(d) g12 = \(\frac{R_2}{R_1+R_2}\), g22 = -R3 – \(\frac{R_1 R_2}{R_1+R_2}\)

The question was posed to me during an interview.

Asked question is from Advanced Problems Involving Parameters in division Two-Port Networks of Network Theory

Answer 1

Right answer is (a) g12 = –\(\frac{R_2}{R_1+R_2}\), g22 = R3 + \(\frac{R_1 R_2}{R_1+R_2}\)

The explanation is: I1 = –\(\frac{R_2}{R_1+R_2}\)I2

Or, g12 = \(\frac{I_1}{I_2} = -\frac{R_2}{R_1+R_2}\)

Also, I2 (R3 + R1 //R2)

= I2 \((R_3 + \frac{R_1 R_2}{R_1+R_2})\)

∴ g22 = \(\frac{V_2}{I_2} = R_3 + \frac{R_1 R_2}{R_1+R_2}\).