Let's solve both circuits using Kirchhoff's Voltage Law (KVL), which states that the sum of all voltages around any closed loop in a circuit is zero.

First Circuit (Left Side):

The circuit consists of two loops and a common resistor.

• Resistors: 100 Ω, 20 Ω, and 10 Ω.
• Voltage Sources: 15 V and 13 V.

Assigning Currents:

• Let $I_1$ be the current in the left loop through the 100 Ω resistor.
• Let $I_2$ be the current in the right loop through the 20 Ω resistor.
• The current through the 10 Ω resistor will be $I_1 - I_2$ since the resistor is shared between the two loops.

Loop Equations:

For the left loop: $15V - 100Ω \cdot I_1 - 10Ω \cdot (I_1 - I_2) = 0$ This simplifies to: $15 - 100I_1 - 10(I_1 - I_2) = 0 \quad \text{(Equation 1)}$ Expanding: $15 - 100I_1 - 10I_1 + 10I_2 = 0$ $15 = 110I_1 - 10I_2 \quad \text{(Equation 1 simplified)}$

For the right loop: $13V - 20Ω \cdot I_2 - 10Ω \cdot (I_2 - I_1) = 0$ This simplifies to: $13 - 20I_2 - 10(I_2 - I_1) = 0 \quad \text{(Equation 2)}$ Expanding: $13 - 20I_2 - 10I_2 + 10I_1 = 0$ $13 = 30I_2 - 10I_1 \quad \text{(Equation 2 simplified)}$

Solving the system of equations:

We now have two equations:

1. $15 = 110I_1 - 10I_2$
2. $13 = 30I_2 - 10I_1$

Let's solve these for $I_1$ and $I_2$.

Second Circuit (Right Side):

This circuit consists of two loops as well.

• Resistors: 3 kΩ, 1.5 kΩ, 1 kΩ, 2 kΩ, and 500 Ω.
• Voltage Sources: 100 V and 130 V.

Assigning Currents:

• Let $I_1$ be the current in the left loop through the 3 kΩ resistor.
• Let $I_2$ be the current in the right loop through the 1.5 kΩ resistor.
• The current through the 1 kΩ resistor will be $I_1 - I_2$ since the resistor is shared between the two loops.

Loop Equations:

For the left loop: $100V - 3kΩ \cdot I_1 - 1kΩ \cdot (I_1 - I_2) - 2kΩ \cdot I_1 = 0$ This simplifies to: $100 - 3000I_1 - 1000(I_1 - I_2) - 2000I_1 = 0 \quad \text{(Equation 1)}$ Expanding: $100 = 6000I_1 - 1000I_2 \quad \text{(Equation 1 simplified)}$

For the right loop: $130V - 1.5kΩ \cdot I_2 - 1kΩ \cdot (I_2 - I_1) - 500Ω \cdot I_2 = 0$ This simplifies to: $130 - 1500I_2 - 1000(I_2 - I_1) - 500I_2 = 0 \quad \text{(Equation 2)}$ Expanding: $130 = 3000I_2 - 1000I_1 \quad \text{(Equation 2 simplified)}$

Solving the system of equations:

We now have two equations:

1. $100 = 6000I_1 - 1000I_2$
2. $130 = 3000I_2 - 1000I_1$

Let me calculate the current values for both circuits.### Solutions for the circuits:

First Circuit (Left Side):

• Current $I_1 = \frac{29}{160} \, \text{A} \approx 0.18125 \, \text{A}$
• Current $I_2 = \frac{79}{160} \, \text{A} \approx 0.49375 \, \text{A}$

Second Circuit (Right Side):

• Current $I_1 = \frac{43}{1700} \, \text{A} \approx 0.02529 \, \text{A}$
• Current $I_2 = \frac{22}{425} \, \text{A} \approx 0.05176 \, \text{A}$

These are the currents in all the branches of the network using KVL analysis.

Would you like a detailed breakdown of how these values were calculated, or do you have any further questions?

Here are 5 follow-up questions:

1. How is Kirchhoff's Voltage Law applied in more complex circuits?
2. What happens if there are dependent sources in the circuit?
3. How does Kirchhoff's Current Law (KCL) relate to KVL?
4. Can these methods be used in AC circuits, and if so, how?
5. How can mesh analysis simplify solving for larger networks?

Tip: When solving complex circuits, labeling the currents and directions carefully helps avoid confusion while writing equations.