Find I1 In The The Network In The Figure

How to Find (i_1) in the Network Shown in the Figure

Finding a specific branch current—such as (i_1)—in an electric circuit is a fundamental skill for students of electrical engineering, physics, and related disciplines. Although the exact topology of the network varies from problem to problem, the underlying methodology remains consistent: apply circuit‑analysis laws systematically, reduce complexity where possible, and solve the resulting algebraic equations. The following guide walks you through a complete, step‑by‑step procedure that you can adapt to any schematic, ensuring you can reliably determine (i_1) even when the figure includes multiple sources, resistors, and possibly dependent elements.

1. Introduction

When presented with a circuit diagram and asked to “find (i_1) in the the network in the figure,” the first task is to interpret the drawing correctly. Identify the branch labeled (i_1), note its direction (the arrow indicates the assumed positive direction), and list all components that share nodes with that branch. Once the circuit is clearly understood, you can choose an analysis technique—mesh analysis, nodal analysis, superposition, or source transformation—that best fits the network’s structure. The goal is to derive a set of independent equations whose solution yields the numerical value of (i_1).

2. Understanding the Circuit

Before jumping into equations, spend a moment extracting the essential information from the figure:

Create a neat schematic (on paper or in your notes) that highlights the branch of interest and labels every node with a number or letter. This visual aid prevents sign errors later on.

3. Choosing an Analysis Method

3.1 Mesh Analysis (Loop Currents)

Mesh analysis is ideal when the circuit contains many voltage sources and relatively few nodes. Steps:

1. Identify meshes – Windows that do not contain any other windows inside them. Assign a mesh current (e.g., (I_A, I_B, …)) to each, following a consistent direction (usually clockwise).
2. Write KVL for each mesh – Sum voltage drops ((IR)) and rises (source voltages) around the loop, setting the sum to zero.
3. Express branch currents – Relate the desired branch current (i_1) to the mesh currents (e.g., if (i_1) flows through a resistor shared by two meshes, (i_1 = I_A - I_B)).
4. Solve the linear system – Use substitution, matrix methods, or a calculator to find the mesh currents, then compute (i_1).

3.2 Nodal Analysis (Node Voltages)

Nodal analysis shines when the circuit has many current sources or when you prefer working with potentials rather than loop currents. Steps:

1. Select a reference node – Ground (0 V). Label all other nodes with unknown voltages ((V_1, V_2, …)).
2. Apply KCL at each non‑reference node – Sum currents leaving the node (using Ohm’s law (I = (V_{node} - V_{adjacent})/R)) and set equal to any injected current sources. 3. Include the branch of interest – If (i_1) flows through a resistor between nodes (V_x) and (V_y), then (i_1 = (V_x - V_y)/R).
3. Solve for node voltages – Obtain (V_x) and (V_y), then calculate (i_1).

3.3 Superposition

When the network contains multiple independent sources, superposition can simplify the problem:

1. Turn off all sources except one – Replace voltage sources with short circuits and current sources with open circuits.
2. Find the contribution to (i_1) from that active source (using mesh or nodal analysis).
3. Repeat for each source and algebraically sum the contributions.
4. Dependent sources remain active throughout; they are not turned off.

3.4 Source Transformation & Thevenin/Norton Equivalents

If a portion of the circuit is “buried” behind the branch of interest, you may replace that portion with its Thevenin or Norton equivalent, reducing the network to a simple series or parallel form where (i_1) is obvious.

4. Step‑by‑Step Example (Illustrative)

Although the actual figure is not provided, we can demonstrate the process with a typical network: a voltage source (V_s = 12\text{ V}), three resistors (R_1 = 2\Omega), (R_2 = 4\Omega), (R_3 = 6\Omega), and a current source (I_s = 1\text{ A}) arranged as shown below (imagine the figure). The branch of interest (i_1) flows through (R_2) from left to right.

4.1 Mesh Analysis Approach

1. Define meshes

   • Mesh A (left loop): includes (V_s), (R_1), and (R_2).
   • Mesh B (right loop): includes (R_2), (R_3), and the current source (I_s).

2. Assign mesh currents (I_A) and (I_B) (both clockwise).

3. Write KVL

   • Mesh A: (-V_s + I_A R_1 + (I_A - I_B)R_2 = 0)
   • Mesh B: ((I_B - I_A)R_2 + I_B R_3 - I_s R_2 = 0) (note the current source contributes a known voltage drop across (R_2) when expressed in mesh form).

4. Insert numbers

   • Mesh A: (-12 + 2I_A + 4(I_A - I_B) = 0 ;\Rightarrow; -12 + 6I_A - 4I_B = 0)
   • Mesh B: (4(I_B - I_A) + 6I_B - 4(1) = 0 ;\Rightarrow; -4I_A + 10I_B - 4 = 0)

5. Solve

   From Mesh A: (6I_A - 4I_B = 12) → (3I_A - 2I_B =