---
title: "Y and g parameters with dependent source 4V1"
subject: "Electrical Circuit Theory"
chapter: 7
key: "semester_3/electrical_circuit_theory:y-and-g-parameters-1h-2f"
source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/y-and-g-parameters-1h-2f
site: BCT Engineering
---

# Y and g parameters with dependent source 4V1

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

- Asked in: 2074 Asoj, 8 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/y-and-g-parameters-1h-2f
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7

## Question

Find the y and g-parameters of the circuit in figure below and also find whether the network is reciprocal or not.
[Figure: port 1 (V1, I1) followed by a 1 H inductor and a 2 F capacitor in series arm; a 3 H inductor as a shunt at port 1 side; a dependent voltage source 4V1 as a middle shunt branch; a 3 H inductor as a shunt at port 2 (V2)]

## Short answer

y = [−8/(3s) 0; −8s (6s²+1)/(3s)]. g-parameters: g11 = −8/(3s), g12 = 0, g21 = 24s²/(6s²+1), g22 = 3s/(6s²+1). The network is non-reciprocal because y12 = 0 and y21 ≠ 0.

## Answer

**Reading the figure (assumption).** From port 1: a 3 H shunt inductor, a 1 H series inductor, then a middle shunt branch that is the dependent voltage source 4V1 (+ at the top), then a 2 F series capacitor, and a 3 H shunt inductor across port 2.

The middle node M is held at VM = 4V1 by the voltage source.

#### y-parameters

```text
Port 1:  I1 = V1/(3s) + (V1 − VM)/s = V1/(3s) + (V1 − 4V1)/s = (1/(3s) − 3/s) V1
         I1 = −(8/(3s)) V1
Port 2:  the current into port 2 node from M through the 2 F is (4V1 − V2)(2s).
         I2 + (4V1 − V2)(2s) = V2/(3s)
         I2 = −8s V1 + (2s + 1/(3s)) V2
```

```text
y11 = −8/(3s)       y12 = 0
y21 = −8s           y22 = 2s + 1/(3s) = (6s² + 1)/(3s)
```

#### g-parameters (I1 = g11 V1 + g12 I2,  V2 = g21 V1 + g22 I2)

Solve the second y-equation for V2 and substitute:

```text
g11 = I1/V1 |I2=0 = y11 − y12 y21/y22 = −8/(3s)
g12 = I1/I2 |V1=0 = y12/y22 = 0
g21 = V2/V1 |I2=0 = −y21/y22 = 8s·3s/(6s² + 1) = 24s²/(6s² + 1)
g22 = V2/I2 |V1=0 = 1/y22 = 3s/(6s² + 1)
```

#### Reciprocity

A network is reciprocal if y12 = y21. Here y12 = 0 but y21 = −8s ≠ 0, so the network is **not reciprocal** (the controlled source makes it active/non-reciprocal). Equivalently g12 = 0 ≠ −g21.

**Answer: y = [−8/(3s) 0; −8s (6s²+1)/(3s)]; g = [−8/(3s) 0; 24s²/(6s²+1) 3s/(6s²+1)]; not reciprocal.**

## Check yourself

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

1. What fixes the voltage of the middle node M?
   A. The 2 F capacitor
   B. The dependent voltage source 4V1, so VM = 4V1
   C. The 3 H inductor at port 2
   D. The current I2
   Answer: B. The middle shunt branch is an ideal voltage source, which holds VM = 4V1. (Key point: Middle node voltage)

2. What are y11 and y12 found in the answer?
   A. y11 = -8/(3s), y12 = 0
   B. y11 = 8/(3s), y12 = -8s
   C. y11 = 1/(3s), y12 = 1/s
   D. y11 = -8s, y12 = 0
   Answer: A. I1 = V1/(3s) + (V1 - 4V1)/s = -(8/(3s))V1, independent of V2. (Key point: y11 and y12)

3. How is g21 obtained from the y-parameters?
   A. g21 = y21/y22
   B. g21 = -y21/y22 = 24s^2/(6s^2+1)
   C. g21 = 1/y22
   D. g21 = y12/y22
   Answer: B. With I2 = 0, V2 = -(y21/y22) V1, giving 24s^2/(6s^2+1). (Key point: g-parameter formula)

4. Why is the network not reciprocal?
   A. y12 = 0 but y21 = -8s, so y12 differs from y21
   B. y11 differs from y22
   C. The determinant of Y is zero
   D. It contains reactive elements
   Answer: A. Reciprocity needs y12 = y21; the controlled source breaks this. (Key point: Reciprocity test)

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