Question

For the circuit given below, the value of the z22 parameter is ___________

(a) z22 = 1 Ω

(b) z22 = 4 Ω

(c) z22 = 1.667 Ω

(d) z22 = 2.33 Ω

Answer 1

Right answer is (c) z22 = 1.667 Ω

For explanation: z11 = \(\frac{V_1}{I_1}\)  = 1 + 6 || (4+2) = 4Ω

I0 =  \(\frac{1}{2} I_1\)

V2 = 2I0 = I1

z21 = \(\frac{V_2}{I_1}\) = 1Ω

z22 = \(\frac{V_2}{I_2}\)  = 2 || (4+6) = 1.667Ω

So, I’0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2\)

V1 = 6I’0 = I2

z12 = \(\frac{V_1}{I_2}\)  = 1Ω

Hence, [z] = [4:1; 1:1.667] Ω.