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 $I_3$.

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