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Let's solve the problem step-by-step using Kirchhoff’s laws as requested.

### Problem Breakdown
The circuit consists of:
- Three resistors with values of \( 1 \, \Omega \), \( 2 \, \Omega \), \( 4 \, \Omega \), \( 8 \, \Omega \), and \( 10 \, \Omega \).
- Three voltage sources of \( 5 \, \text{V} \), \( 20 \, \text{V} \), and \( 10 \, \text{V} \).
- Three currents \( I_1 \), \( I_2 \), and \( I_3 \) as indicated in the diagram.

### Given Questions
1. **(a)** Write down the current equation at node \( b \) in terms of \( I_1 \), \( I_2 \), and \( I_3 \).
2. **(b)** Write an equation for the circuit path "abefa" in terms of \( I_1 \) and \( I_2 \).
3. **(c)** Write an equation for the circuit path "cbedc" in terms of \( I_1 \) and \( I_2 \).
4. **(d)** Determine the values of \( I_1 \), \( I_2 \), and \( I_3 \).

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### Step-by-Step Solution

#### Step 1: Apply Kirchhoff’s Current Law (KCL) at Node \( b \)
At node \( b \), the sum of currents flowing into the node must equal the sum of currents flowing out. From the diagram:
\[
I_1 = I_2 + I_3
\]
This is the current equation at node \( b \).

#### Step 2: Apply Kirchhoff’s Voltage Law (KVL) to Path "abefa"
For the loop "abefa":
- Starting from \( a \) and moving to \( b \), we pass through the \( 1 \, \Omega \) resistor. The voltage drop is \( 1 \cdot I_1 \).
- Moving from \( b \) to \( e \), we pass through the \( 10 \, \Omega \) resistor. The voltage drop is \( 10 \cdot I_2 \).
- Moving from \( e \) to \( f \), we pass through the \( 4 \, \Omega \) resistor. The voltage drop is \( 4 \cdot I_1 \).
- Moving from \( f \) back to \( a \), we pass through the \( 5 \, \text{V} \) voltage source.

Applying KVL:
\[
-5 + (1 \cdot I_1) + (10 \cdot I_2) + (4 \cdot I_1) = 0
\]
Simplify this equation:
\[
5 I_1 + 10 I_2 = 5 \quad \Rightarrow \quad I_1 + 2 I_2 = 1 \tag{Equation 1}
\]

#### Step 3: Apply KVL to Path "cbedc"
For the loop "cbedc":
- Starting from \( c \) and moving to \( b \), we pass through the \( 2 \, \Omega \) resistor. The voltage drop is \( 2 \cdot I_3 \).
- Moving from \( b \) to \( e \), we pass through the \( 10 \, \Omega \) resistor. The voltage drop is \( 10 \cdot I_2 \).
- Moving from \( e \) to \( d \), we pass through the \( 8 \, \Omega \) resistor. The voltage drop is \( 8 \cdot I_3 \).
- Moving from \( d \) back to \( c \), we pass through the \( 10 \, \text{V} \) voltage source.

Applying KVL:
\[
-10 + (2 \cdot I_3) + (10 \cdot I_2) + (8 \cdot I_3) = 0
\]
Simplify this equation:
\[
10 I_3 + 10 I_2 = 10 \quad \Rightarrow \quad I_3 + I_2 = 1 \tag{Equation 2}
\]

#### Step 4: Solve the System of Equations
We now have three equations:
1. \( I_1 = I_2 + I_3 \)
2. \( I_1 + 2 I_2 = 1 \)
3. \( I_3 + I_2 = 1 \)

Using substitution, we can solve these equations:

From Equation 3:
\[
I_3 = 1 - I_2
\]

Substitute \( I_3 = 1 - I_2 \) into Equation 1:
\[
I_1 = I_2 + (1 - I_2) = 1
\]

Now that we know \( I_1 = 1 \), substitute \( I_1 = 1 \) into Equation 2:
\[
1 + 2 I_2 = 1 \quad \Rightarrow \quad

Refer to the circuit shown in Figure 1:
(a) Write down the current equation at node b in terms of I1, I2, and I3.
(b) Write an equation for the circuit path 'abefa' in terms of I1 and I2.
(c) Write an equation for the circuit path 'cbedc' in terms of I1 and I2.
(d) Find the values of I1, I2, and I3.

Solve a circuit analysis problem using Kirchhoff's Current Law and Kirchhoff's Voltage Law to find currents I1, I2, and I3 in a circuit with resistors and voltage sources. Detailed solution suitable for Electrical Engineering students.

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