Series Resistor Calculation

This explanation details how to determine the resistance of a second resistor when connected in series with an initial resistor to halve the circuit's current under constant voltage.

Initial Circuit Analysis

Let the initial resistor be $R_1 = 10 \text{ Ω}$. The circuit is connected to a constant voltage source $V$. According to Ohm's Law, the initial current $I_1$ is:

$I_1 = \frac{V}{R_1}$

Series Circuit Setup

A second resistor, denoted as $R_2$, is added in series with $R_1$. The total resistance $R_{total}$ in the circuit becomes the sum of the two resistances:

$R_{total} = R_1 + R_2$

The new current, $I_2$, flowing through the series combination is:

$I_2 = \frac{V}{R_1 + R_2}$

Current Relationship

The problem states that the new current $I_2$ is exactly half of the initial current $I_1$:

$I_2 = \frac{I_1}{2}$

Determining the Second Resistor's Value

Substitute the expressions for $I_1$ and $I_2$ into the current relationship equation:

$\frac{V}{R_1 + R_2} = \frac{1}{2} \times \frac{V}{R_1}$

Since the voltage $V$ remains constant, it can be canceled from both sides of the equation:

$\frac{1}{R_1 + R_2} = \frac{1}{2R_1}$

By cross-multiplying:

$2R_1 = R_1 + R_2$

To find $R_2$, rearrange the equation:

$R_2 = 2R_1 - R_1$

$R_2 = R_1$

Final Result

Given that the initial resistor $R_1 = 10 \text{ Ω}$, the value of the second resistor $R_2$ must be equal to $R_1$:

$R_2 = 10 \text{ Ω}$

Thus, the resistance of the second resistor is 10 Ω.