To find the emf $E_1$ in the middle branch, we need to use Kirchhoff's circuit laws, specifically the loop rule and the junction rule. Let's break the problem down step-by-step:

Step 1: Analyze the Given Information

We have three parallel branches:

1. Top branch:

   • Resistors: $1.00 \, \Omega$ and $6.00 \, \Omega$ in series
   • Battery: $20.0 \, \text{V}$ with a current of $1.00 \, \text{A}$ flowing from right to left

2. Middle branch:

   • Resistors: $4.00 \, \Omega$ and $1.00 \, \Omega$ in series
   • Battery: $E_1$ (unknown emf)
   • The current in the middle branch is unknown.

3. Bottom branch:

   • Resistor: $1.00 \, \Omega$
   • Battery: $E_2$ (unknown emf)
   • Resistor: $2.00 \, \Omega$
   • The current in this branch is $2.00 \, \text{A}$ from left to right.

Step 2: Apply Kirchhoff's Current Law (KCL)

At node $a$, the total current entering or leaving must be conserved.

• The current from the top branch is $1.00 \, \text{A}$ (right to left).
• The current from the bottom branch is $2.00 \, \text{A}$ (left to right).
• Therefore, the current through the middle branch, which we'll call $I_1$, must be: $I_1 = 3.00 \, \text{A} - 2.00 \, \text{A} = 1.00 \, \text{A}$

Step 3: Apply Kirchhoff's Voltage Law (KVL) for the Top Branch

In the top branch, there are two resistors and a battery in series. The total voltage drop across the resistors and the battery should sum to zero when we go around the loop.

• The total resistance in the top branch is $1.00 \, \Omega + 6.00 \, \Omega = 7.00 \, \Omega$.
• The current in the top branch is $1.00 \, \text{A}$, so the voltage drop across the resistors is: $V_{\text{resistors}} = I \times R = 1.00 \, \text{A} \times 7.00 \, \Omega = 7.00 \, \text{V}$
• The total voltage from the battery is $20.0 \, \text{V}$.

Using KVL, we write the equation for the top branch: $V_{\text{battery}} - V_{\text{resistors}} = 0$ $20.0 \, \text{V} - 7.00 \, \text{V} = 13.00 \, \text{V}$ This is the potential difference across the top branch.

Step 4: Apply KVL for the Middle Branch

In the middle branch, there are two resistors and the battery with emf $E_1$.

• The total resistance in the middle branch is $4.00 \, \Omega + 1.00 \, \Omega = 5.00 \, \Omega$.
• The current in this branch is $I_1 = 1.00 \, \text{A}$.
• The voltage drop across the resistors is: $V_{\text{resistors}} = I_1 \times R = 1.00 \, \text{A} \times 5.00 \, \Omega = 5.00 \, \text{V}$
• The potential difference across the battery in the middle branch is $E_1$.

Now, apply KVL for the middle branch: $E_1 - V_{\text{resistors}} = 0$ $E_1 - 5.00 \, \text{V} = 0$ Thus, we find: $E_1 = 5.00 \, \text{V}$

Conclusion:

The emf $E_1$ in the middle branch is 5.00 V.

---

Would you like to go over the detailed steps again, or is there anything else you'd like clarification on? Here are some related questions you could explore further:

1. How do you solve for the emf in a circuit with multiple resistors in series and parallel?
2. How does Kirchhoff's Current Law apply in multi-branch circuits?
3. What would happen if one of the resistors in the top branch had a different value?
4. How would the current in the middle branch change if $E_1$ were higher?
5. How can you calculate the total current flowing through a parallel circuit?

Tip: Always check your current directions and voltage signs when applying Kirchhoff's laws, as they help ensure your calculations are accurate!