Understanding Resistance Change in a Stretched Wire

This question involves understanding how the electrical resistance of a wire changes when its physical dimensions are altered, specifically when it is stretched. We also need to apply Ohm's Law to determine the effect on the current in a circuit when the resistance changes, assuming the voltage remains constant.

How Stretching Affects Resistance

The resistance ($R$) of a wire is directly proportional to its length ($L$) and inversely proportional to its cross-sectional area ($A$). The formula is given by:

$$R = \rho \frac{L}{A}$$

where $\rho$ is the resistivity of the material, which remains constant for a given wire material at a constant temperature.

When a wire is stretched, its length increases. Assuming the volume of the wire remains constant (which is typical for stretching), its cross-sectional area must decrease proportionally.

Let the original length be $L_1$ and the original cross-sectional area be $A_1$. The original resistance is:

$$R_1 = \rho \frac{L_1}{A_1}$$

The wire is stretched to a new length $L_2 = 2L_1$. Since the volume $V = A \times L$ is constant, we have:

$$V_1 = V_2$$

$$A_1 L_1 = A_2 L_2$$

Substituting $L_2 = 2L_1$:

$$A_1 L_1 = A_2 (2L_1)$$

Dividing both sides by $L_1$ (assuming $L_1 \neq 0$), we get the new area $A_2$:

$$A_2 = \frac{A_1 L_1}{2L_1} = \frac{A_1}{2}$$

So, when the length is doubled by stretching, the cross-sectional area is halved.

Calculating the New Resistance

Now we can find the new resistance $R_2$ using the new length $L_2 = 2L_1$ and the new area $A_2 = A_1/2$:

$$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1/2}$$

$$R_2 = \rho \frac{2L_1 \times 2}{A_1} = \rho \frac{4L_1}{A_1}$$

We can rewrite this in terms of the original resistance $R_1 = \rho \frac{L_1}{A_1}$:

$$R_2 = 4 \times \left( \rho \frac{L_1}{A_1} \right) = 4 R_1$$

The new resistance is four times the original resistance. Given the original resistance is 10 Ω:

$$R_2 = 4 \times 10 \, \Omega = 40 \, \Omega$$

The resistance of the wire increases from 10 Ω to 40 Ω after stretching.

Impact on Current using Ohm's Law

Ohm's Law states that the current ($I$) flowing through a conductor is directly proportional to the voltage ($V$) across it and inversely proportional to its resistance ($R$), provided the temperature and other physical conditions remain unchanged. The relationship is $V = IR$, or rearranged to find current:

$$I = \frac{V}{R}$$

In this circuit, we assume the voltage source connected across the wire remains constant. Let the constant voltage be $V$.

The original current $I_1$ with resistance $R_1$ is:

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

The new current $I_2$ with resistance $R_2$ is:

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

We found that $R_2 = 4R_1$. Substituting this into the equation for $I_2$:

$$I_2 = \frac{V}{4R_1}$$

We can factor out the $\frac{1}{4}$:

$$I_2 = \frac{1}{4} \times \frac{V}{R_1}$$

Since $I_1 = \frac{V}{R_1}$, we can write:

$$I_2 = \frac{1}{4} I_1$$

The new current is one-fourth of the original current.

Summary of Changes

• Original Resistance $R_1 = 10 \, \Omega$.
• Wire stretched to double length $L_2 = 2L_1$.
• Volume remains constant, so Area changes to $A_2 = A_1/2$.
• New Resistance $R_2 = 4R_1 = 40 \, \Omega$.
• Assuming constant voltage $V$.
• Original Current $I_1 = V/R_1$.
• New Current $I_2 = V/R_2 = V/(4R_1) = (1/4)(V/R_1) = (1/4)I_1$.

Therefore, the current in the circuit would become one-fourth of its original value.

Revision Table: Resistance and Current Formulas

Additional Information: Related Concepts

Resistivity ($\rho$): This is an intrinsic property of the material the wire is made of. It indicates how strongly the material resists the flow of electric current. Different materials have different resistivities (e.g., copper has low resistivity, rubber has very high resistivity).

Conductivity ($\sigma$): This is the inverse of resistivity ($\sigma = 1/\rho$). It measures how easily electric current flows through a material. Good conductors have high conductivity and low resistivity.

Effect of Temperature: For most conductors (like metals), resistance increases with increasing temperature. The resistivity $\rho$ is temperature-dependent. In this problem, we assumed the temperature remains constant, so $\rho$ is constant.

Power Dissipation: When current flows through a resistor, electrical energy is converted into heat. The power dissipated ($P$) is given by $P = VI = I^2R = V^2/R$. If the resistance quadruples and the voltage is constant, the power dissipated becomes $P_2 = V^2 / R_2 = V^2 / (4R_1) = (1/4)P_1$. If the resistance quadruples and the current is constant (which is not the case here as current changes), the power would be $P_2 = I^2 R_2 = I^2 (4R_1) = 4P_1$.