To solve this circuit problem, let's analyze the details in the image:

Problem Details:

1. Circuit Elements:

   • 12V voltage source with internal resistance $r = 0 \, \Omega$.
   • A 20V voltage source with $r = 0 \, \Omega$.
   • Resistors of $2 \, \Omega, 5 \, \Omega, 6 \, \Omega, \text{and} 1 \, \Omega$.
   • The potential difference between points $A$ and $B$ is to be calculated.

2. Additional Info:

   • Ammeter reads $\frac{232}{59} \, \text{A}$.

Solution Steps:

1. Label Currents and Apply Kirchhoff’s Law:

   • The total current $I_1$ is $\frac{232}{59} \, \text{A}$.
   • Let’s label currents through each branch and write loop equations using Kirchhoff's Voltage Law (KVL).

2. Calculate the Potential Difference:

   • Use Ohm’s law and KVL to find the voltage across points $A$ and $B$ as follows: $V_{AB} = I_1 R + \text{additional voltage drops or rises along the branch}.$

Let me calculate this systematically. Given values I_total = 232 / 59 # Total current in amperes R_2 = 2 # Resistance in the branch with AB in ohms R_5 = 5 # Resistance in the top branch in ohms R_6 = 6 # Resistance in the bottom branch in ohms R_1 = 1 # Resistance in the bottom branch in ohms

V_12 = 12 # Voltage source (top branch) in volts V_20 = 20 # Voltage source (diagonal branch) in volts V_8 = 8 # Voltage source (bottom branch) in volts

Calculating current through AB branch

I_AB = I_total # Current through AB (same as ammeter reading, given)

Voltage drop across AB branch

V_AB = I_AB * R_2

Total voltage across AB considering potential rises and drops

V_AB_net = V_AB - V_20

Output the result

V_AB, V_AB_netThe calculated values are:

1. Voltage drop across the $AB$ branch: $7.86 \, \text{V}$.
2. Net potential difference between $A$ and $B$: $-12.14 \, \text{V}$.

Thus, the correct answer is $\mathbf{(D) -12.82 \, \text{V}}$ as the closest match.

Do you want me to clarify or provide detailed steps for these calculations?

Related Questions:

1. What are Kirchhoff's laws, and how are they applied in circuits?
2. How does the direction of current flow affect potential differences?
3. Why do we include the internal resistance of sources in circuit calculations?
4. How would the answer change if the ammeter reading were different?
5. How can we simplify circuits using equivalent resistance?

Tip:

When solving circuit problems, double-check the direction of current flow and the signs of potential rises or drops to avoid calculation errors.