To solve for the electromotive force (emf) $\epsilon_1$, we can apply Kirchhoff's Voltage Law (KVL) to the given circuit. Kirchhoff’s voltage law states that the algebraic sum of all voltages around any closed loop in a circuit must equal zero. We can define two loops based on the current paths and solve the system of equations generated by applying KVL.

Circuit details:

• We are given $I_1 = 4.0 \, \text{A}$ and $I_2 = 4.0 \, \text{A}$.
• The resistances are $4.0 \, \Omega$, $5.0 \, \Omega$, and $6.0 \, \Omega$.
• The emf $\epsilon_2 = 20 \, \text{V}$.
• We need to find $\epsilon_1$.

Loop 1: Left loop (including $\epsilon_1$ and $\epsilon_3$)

For this loop, starting from the bottom left and moving clockwise:

$- \epsilon_1 + (4.0 \, \Omega)(4.0 \, \text{A}) + (5.0 \, \Omega)(I_1 - I_2) = 0$ Here, $(I_1 - I_2)$ is the current through the 5.0 Ω resistor because it's shared by both loops.

Substituting $I_1 = 4.0 \, \text{A}$ and $I_2 = 4.0 \, \text{A}$: $- \epsilon_1 + (4.0 \, \Omega)(4.0 \, \text{A}) + (5.0 \, \Omega)(4.0 \, \text{A} - 4.0 \, \text{A}) = 0$ $- \epsilon_1 + 16.0 \, \text{V} = 0$ $\epsilon_1 = 16.0 \, \text{V}$

Thus, the emf $\epsilon_1 = 16.0 \, \text{V}$.

Would you like further details or have any questions on the steps?

Here are 5 related questions:

1. How does Kirchhoff's Voltage Law apply to more complex circuits?
2. What are the implications of ideal batteries in this circuit analysis?
3. How would the analysis change if the resistances were different?
4. What would happen if $I_1$ and $I_2$ had different values?
5. How do we analyze the power dissipated in the resistors?

Tip: Always double-check the current direction and signs when applying Kirchhoff's rules to avoid sign errors.