When calculated from the balanced length, the potentiometer difference across the first resistor of 58 is 2.6V. The current through the second resistor from 582 to the equilibrium point is:

In this case, we can apply Ohm's Law and the principles of a potentiometer to find the current through the second resistor. Since the question involves a potentiometer, the "balanced length" refers to the length where the voltage across the circuit is balanced, meaning there is no current through the galvanometer in the potentiometer setup.

### Given:
- Voltage difference across the first resistor = 2.6V
- First resistor value, \( R_1 = 58 \, \Omega \)
- Second resistor value, \( R_2 = 582 \, \Omega \)

To find the current through the second resistor, we first calculate the total current in the circuit. By Ohm's law, the voltage across the first resistor can be used to calculate the current through it:

\[
I = \frac{V_1}{R_1} = \frac{2.6 \, \text{V}}{58 \, \Omega}
\]

Let's calculate this.The current through the first resistor, and hence through the entire circuit, is approximately \( 0.0448 \, \text{A} \) or 44.8 mA.

Since the same current flows through both resistors in series, the current through the second resistor \( R_2 = 582 \, \Omega \) will also be \( 0.0448 \, \text{A} \).

Would you like further details, or do you have any questions?

Here are 5 related questions to expand on this:

1. How does a potentiometer measure voltage in a circuit?
2. What is the significance of the balanced point in a potentiometer?
3. How would the current change if the value of the second resistor \(R_2\) were doubled?
4. What is the total resistance of the circuit with both resistors in series?
5. Can we calculate the total voltage drop across both resistors?

**Tip**: Always ensure the units are consistent when applying Ohm's law (e.g., volts, ohms, amperes) to avoid calculation errors.

This example demonstrates how to calculate the current through a resistor in a series circuit using Ohm's Law. The problem involves a potentiometer with a 58-ohm resistor and a 582-ohm resistor. Suitable for students in Grades 10-12, this solution explains key concepts of voltage, current, and resistance in series circuits.

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