---
title: "Z and T parameters with CCCS 4V1"
subject: "Electrical Circuit Theory"
chapter: 7
key: "semester_3/electrical_circuit_theory:z-and-t-parameters-4v1-2ohm"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/z-and-t-parameters-4v1-2ohm
site: BCT Engineering
---

# Z and T parameters with CCCS 4V1

Electrical Circuit Theory · Chapter 7: Two-port Parameter of Networks

- Asked in: 2072 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/z-and-t-parameters-4v1-2ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7

## Question

Find Z-parameters and hence the T-parameters for the 2 port network shown in figure below.
[Figure: port 1 (V1, I1); a 2 Ω shunt resistor at port 1; a 1 Ω series resistor and a 2 Ω series resistor in the top wire; a dependent current source 4V1 (arrow pointing down) as a shunt at the junction of the two series resistors; a 1 Ω shunt resistor at port 2 (V2, I2)]

## Short answer

Form node equations to obtain Y parameters, then invert to Z: z11 = 4/15 Ω, z12 = 1/15 Ω, z21 = −1/5 Ω, z22 = 7/10 Ω. Convert to T: A = z11/z21 = −4/3, B = Δz/z21 = −1 Ω, C = 1/z21 = −5 S, D = z22/z21 = −7/2.

## Answer

Z-parameters: V1 = z11 I1 + z12 I2, V2 = z21 I1 + z22 I2. Transmission parameters: V1 = A V2 − B I2, I1 = C V2 − D I2.

**Circuit.** 2 Ω shunt at port 1; 1 Ω then 2 Ω in the series arm; a dependent current source 4V1 (arrow down) in shunt at the junction J of the two series resistors; 1 Ω shunt at port 2.

Because the sources and the series resistors form a ladder, it is easiest to write the node equations and solve for the currents in terms of voltages (Y form), then invert.

#### Node equations

```text
Port 1 node:   I1 = V1/2 + (V1 − VJ)/1
Node J:        (V1 − VJ)/1 = (VJ − V2)/2 + 4V1
Port 2 node:   I2 + (VJ − V2)/2 = V2/1
```

From node J: −3V1 − 1.5 VJ + 0.5 V2 = 0, so VJ = V2/3 − 2V1.

```text
I1 = 1.5 V1 − VJ = 3.5 V1 − V2/3
I2 = 1.5 V2 − VJ/2 = V1 + (4/3) V2
```

Thus y11 = 7/2, y12 = −1/3, y21 = 1, y22 = 4/3, and Δy = (7/2)(4/3) + 1/3 = 5.

#### Z-parameters (Z = Y⁻¹)

```text
z11 = y22/Δy = 4/15 Ω       z12 = −y12/Δy = 1/15 Ω
z21 = −y21/Δy = −1/5 Ω      z22 = y11/Δy = 7/10 Ω
```

#### T-parameters

```text
A = z11/z21 = −4/3          B = Δz/z21 = −1 Ω     (Δz = 1/5)
C = 1/z21   = −5 S          D = z22/z21 = −7/2
Check: AD − BC = 14/3 − 5 = −1/3 ≠ 1 → not reciprocal.
```

**Answer: Z = [4/15 1/15; −1/5 7/10] Ω; T: A = −4/3, B = −1 Ω, C = −5 S, D = −7/2.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What approach does the answer use to find Z?
   A. Open-circuit tests only
   B. Write node equations to get Y, then invert Z = Y^-1
   C. Mesh equations with no inversion
   D. Use the h-parameters
   Answer: B. The ladder with dependent source is easiest to solve in Y form and then invert. (Key point: Method)

2. What does the node J equation give for VJ?
   A. VJ = V2/3 - 2V1
   B. VJ = 2V1 + V2
   C. VJ = V1 - V2
   D. VJ = 4V1
   Answer: A. From -3V1 - 1.5VJ + 0.5V2 = 0, VJ = V2/3 - 2V1. (Key point: Node J result)

3. AD - BC = -1/3. What does this show?
   A. Reciprocal network
   B. Not reciprocal, because it is not 1
   C. Symmetrical network
   D. The network is passive
   Answer: B. Reciprocity requires AD - BC = 1. (Key point: Reciprocity check)

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