This circuit consists of three loops with resistors and two voltage sources. Let's analyze it step-by-step:

1. Labels:

• The circuit has the following components:
  • Voltage Sources:
    • 80 V (between points E and F)
    • 20 V (between points D and F)
  • Resistors:
    • 30 Ω between points A and B
    • 20 Ω between points C and D
    • 20 Ω between points F and D
  • Currents:
    • $I_1$ in the loop A-B-D (clockwise),
    • $I_2$ in the loop E-C-D-F (clockwise),
    • $I_3$ in the middle branch C-D.

2. Kirchhoff’s Laws Application:

To solve for the currents, we can apply Kirchhoff's Voltage Law (KVL) to the loops.

Loop 1: A → B → D → C → A:

$-I_1 \times 30 \, \Omega + 20 \, V - I_3 \times 20 \, \Omega = 0$ Rearranging: $30I_1 - 20 - 20I_3 = 0 \quad \text{(Equation 1)}$

Loop 2: E → C → D → F → E:

$80 \, V - I_2 \times 20 \, \Omega - I_3 \times 20 \, \Omega = 0$ Rearranging: $80 - 20I_2 - 20I_3 = 0 \quad \text{(Equation 2)}$

Middle Branch (C-D):

By Kirchhoff’s current law (KCL), the current $I_3$ in the middle branch is the difference between $I_1$ and $I_2$: $I_3 = I_1 - I_2 \quad \text{(Equation 3)}$

3. Solving the Equations:

Now we have three equations:

1. $30I_1 - 20 - 20I_3 = 0$
2. $80 - 20I_2 - 20I_3 = 0$
3. $I_3 = I_1 - I_2$

We can substitute equation (3) into equations (1) and (2) and solve for $I_1$, $I_2$, and $I_3$.

Would you like me to solve these equations for you and find the values of $I_1$, $I_2$, and $I_3$?

Here are 5 related questions:

1. How would the currents change if the 20 V source was reversed in polarity?
2. What happens to the currents if the resistance of the 30 Ω resistor is increased?
3. Can you apply Thevenin’s Theorem to simplify the circuit?
4. What is the power dissipated in each resistor?
5. How does Kirchhoff’s Current Law apply to point C?

Tip: Always check the direction of the current assumptions when applying Kirchhoff's Laws to ensure consistency.