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The image contains a circuit diagram with three loops and resistances. Using Kirchhoff’s laws (KCL and KVL), we can calculate the currents \(i_1\), \(i_2\), and \(i_3\) flowing through each part of the circuit.

### Circuit elements:
- \(V_1 = 80 \text{V}\), \(V_2 = 90 \text{V}\)
- Resistors: \(20 \Omega\), \(10 \Omega\) (horizontal branch), and \(15 \Omega\)
- Currents: \(i_1\) (left branch), \(i_2\) (middle branch), \(i_3\) (right branch)

### Applying Kirchhoff's Current Law (KCL) at node \(Q\):
At node \(Q\), the sum of currents entering the node equals the sum of currents leaving:
\[
i_1 = i_2 + i_3
\]
This equation will be useful for solving the system.

### Applying Kirchhoff's Voltage Law (KVL):
1. **Loop 1 (left loop with 80 V source):**
   Moving in the clockwise direction:
   \[
   -80V + 20\Omega \cdot i_1 + 10\Omega \cdot i_2 = 0
   \]
   Simplifying:
   \[
   20i_1 + 10i_2 = 80 \quad \text{(Equation 1)}
   \]

2. **Loop 2 (right loop with 90 V source):**
   Moving in the clockwise direction:
   \[
   -90V + 15\Omega \cdot i_3 + 10\Omega \cdot i_2 = 0
   \]
   Simplifying:
   \[
   15i_3 + 10i_2 = 90 \quad \text{(Equation 2)}
   \]

3. **Loop 3 (entire outer loop):**
   Moving clockwise from the left voltage source to the right voltage source:
   \[
   -80V + 20\Omega \cdot i_1 + 15\Omega \cdot i_3 + 90V = 0
   \]
   Simplifying:
   \[
   20i_1 + 15i_3 = -10 \quad \text{(Equation 3)}
   \]

### Solving the system of equations:
We have the following system of equations:
1. \( 20i_1 + 10i_2 = 80 \)
2. \( 15i_3 + 10i_2 = 90 \)
3. \( 20i_1 + 15i_3 = -10 \)

I will now solve these equations to find \(i_1\), \(i_2\), and \(i_3\).The solution to the system of equations gives the following values for the currents:

- \( i_1 = -0.5 \, \text{A} \)
- \( i_2 = 9 \, \text{A} \)
- \( i_3 = 0 \, \text{A} \)

This indicates:
- Current \( i_1 \) flows in the opposite direction as initially assumed.
- Current \( i_2 \) is quite strong in the middle branch.
- There is no current (\( i_3 = 0 \)) flowing in the rightmost branch of the circuit.

Would you like a detailed explanation of these results or further clarification?

Here are 5 related questions you might find helpful:
1. How would the solution change if the rightmost voltage source was different?
2. What happens to the current directions if all resistances are halved?
3. How does Kirchhoff's Current Law apply to complex circuits with multiple junctions?
4. What role do the values of resistances play in determining the current distribution?
5. Can we apply Thevenin’s theorem to simplify this circuit?

**Tip:** When using Kirchhoff's laws, always assume current directions. If the result is negative, it means the current flows in the opposite direction.

Apply Kirchhoff's laws to find the currents in the circuit, given a 20Ω, 10Ω, and 15Ω resistor, and voltage sources of 80V and 90V.

Learn how to apply Kirchhoff's Current and Voltage Laws to find currents in a circuit with resistors of 20Ω, 10Ω, 15Ω, and voltage sources of 80V and 90V. This step-by-step solution involves solving a system of linear equations and analyzing current directions.

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