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To determine the current in each branch using nodal analysis for the given circuit, let's first outline the steps:

### Step 1: Identify Nodes
Let’s assign nodes for clarity:
- Node 1: The top point between the 3Ω, 10Ω, and 5Ω resistors.
- Ground: The bottom of the circuit (reference node).

### Step 2: Apply KCL (Kirchhoff's Current Law) at each node
At each node, the sum of the currents leaving the node is zero.

#### For Node 1:
Let the voltage at Node 1 be \( V_1 \) with respect to the reference node (ground).

Using Ohm’s law for each branch, we express the currents in terms of \( V_1 \):

1. **Current through the 3Ω resistor:**
   \[
   I_{\text{3Ω}} = \frac{V_1}{3}
   \]
   
2. **Current through the 10Ω resistor (connected between Node 1 and the 5A current source):**
   Since there is a 5A current source on the left, the current through the 10Ω resistor must balance with that current source:
   \[
   I_{\text{10Ω}} = \frac{V_1}{10}
   \]

3. **Current through the 5Ω resistor (connected between Node 1 and the 10A current source):**
   This is the current flowing through the 5Ω resistor due to the 10A current source on the right:
   \[
   I_{\text{5Ω}} = \frac{V_1}{5}
   \]

Now, applying KCL at Node 1:

\[
\text{Current entering} = \text{Current leaving}
\]
\[
5A + \frac{V_1}{10} = \frac{V_1}{3} + \frac{V_1}{5} + 10A
\]

### Step 3: Simplify the Equation
To simplify, let’s first multiply through by the least common denominator, which is 30:

\[
30 \times 5 + 30 \times \frac{V_1}{10} = 30 \times \frac{V_1}{3} + 30 \times \frac{V_1}{5} + 30 \times 10
\]

\[
150 + 3V_1 = 10V_1 + 6V_1 + 300
\]

Now combine like terms:

\[
150 + 3V_1 = 16V_1 + 300
\]

\[
150 - 300 = 16V_1 - 3V_1
\]

\[
-150 = 13V_1
\]

Solve for \( V_1 \):

\[
V_1 = \frac{-150}{13} \approx -11.54 \text{ volts}
\]

### Step 4: Calculate the Currents in Each Branch
Now that we have \( V_1 \), we can calculate the currents through each resistor:

1. **Current through the 3Ω resistor:**
   \[
   I_{\text{3Ω}} = \frac{V_1}{3} = \frac{-11.54}{3} \approx -3.85 \text{ A}
   \]

2. **Current through the 10Ω resistor:**
   \[
   I_{\text{10Ω}} = \frac{V_1}{10} = \frac{-11.54}{10} \approx -1.15 \text{ A}
   \]

3. **Current through the 5Ω resistor:**
   \[
   I_{\text{5Ω}} = \frac{V_1}{5} = \frac{-11.54}{5} \approx -2.31 \text{ A}
   \]

### Final Answer:
- Current through 3Ω: \( -3.85 \, \text{A} \) (negative indicates direction opposite to assumed).
- Current through 10Ω: \( -1.15 \, \text{A} \).
- Current through 5Ω: \( -2.31 \, \text{A} \).

Let me know if you'd like a detailed explanation for any of these steps!

### Follow-up Questions:
1. How is the polarity of voltages determined in nodal analysis?
2. What happens if one of the resistances is replaced with a capacitor?
3. How would the current values change if the 5A current source was changed?
4. Can you explain the impact of having more nodes in this circuit?
5. How would mesh analysis differ in solving this circuit?

### Tip:
In nodal analysis, always make sure to correctly set up the ground (reference) and ensure consistency in assuming the direction of currents.

Determine the current in each branch using the Nodal analysis for the given circuit.

Learn how to determine the current in each branch of a circuit using nodal analysis, Kirchhoff's Current Law, and Ohm's Law. This example involves resistors and current sources, suitable for Grades 11-12 students.

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