1.

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 ΩI got this question by my college professor while I was bunking the class.The doubt is from Advanced Problems on Two Port Network topic in chapter Two-Port Networks of Network Theory

Answer»

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.



Discussion

No Comment Found

Related InterviewSolutions