What happens to the ammeter reading when the length of the wire is doubled?
It decreases to one half
The question asks what happens to the ammeter reading when the length of a wire is doubled. An ammeter measures the electric current flowing through a circuit. To understand how changing the wire's length affects the current, we need to consider the concept of electrical resistance.
Electrical resistance (R) is a property of a material that opposes the flow of electric current. The resistance of a uniform wire depends on several factors, including:
The relationship is given by the formula:
\( R = \frac{\rho L}{A} \)
This formula shows that resistance (R) is directly proportional to the length (L) of the wire. If you double the length of the wire while keeping the material (resistivity) and cross-sectional area the same, the resistance will also double.
Ohm's Law relates voltage (V), current (I), and resistance (R) in a circuit. It is stated as:
\( V = IR \)
This can be rearranged to find the current (I), which is what the ammeter measures:
\( I = \frac{V}{R} \)
Assuming the voltage (V) across the wire remains constant, the current (I) is inversely proportional to the resistance (R). This means if the resistance increases, the current decreases, and vice versa.
Let the original length of the wire be \(L_1\) and the original resistance be \(R_1\). When the length is doubled, the new length is \(L_2 = 2L_1\). Since resistance is directly proportional to length:
\( R_1 = \frac{\rho L_1}{A} \)
\( R_2 = \frac{\rho L_2}{A} = \frac{\rho (2L_1)}{A} = 2 \left( \frac{\rho L_1}{A} \right) = 2R_1 \)
So, the new resistance \(R_2\) is double the original resistance \(R_1\).
Now, let the original current be \(I_1\) and the new current be \(I_2\). Using Ohm's Law (assuming voltage V is constant):
\( I_1 = \frac{V}{R_1} \)
\( I_2 = \frac{V}{R_2} \)
Substitute \(R_2 = 2R_1\) into the equation for \(I_2\):
\( I_2 = \frac{V}{2R_1} = \frac{1}{2} \left( \frac{V}{R_1} \right) \)
Since \(I_1 = \frac{V}{R_1}\), we can write:
\( I_2 = \frac{1}{2} I_1 \)
This shows that the new current \(I_2\) is half of the original current \(I_1\). The ammeter reading, which measures the current, will therefore decrease to one half when the length of the wire is doubled.
| Property | Original State | When Length is Doubled | Relationship |
|---|---|---|---|
| Length (L) | \(L\) | \(2L\) | \(L_{new} = 2L_{old}\) |
| Resistance (R) | \(R = \frac{\rho L}{A}\) | \(R' = \frac{\rho (2L)}{A} = 2R\) | \(R_{new} = 2R_{old}\) |
| Current (I) (assuming constant V) | \(I = \frac{V}{R}\) | \(I' = \frac{V}{R'} = \frac{V}{2R} = \frac{1}{2} I\) | \(I_{new} = \frac{1}{2} I_{old}\) |
Based on the analysis using resistance formula and Ohm's Law, doubling the length of the wire doubles its resistance. A constant voltage source then drives half the amount of current through the wire. Therefore, the ammeter reading decreases to one half.
| Change in Wire Property | Effect on Resistance | Effect on Current (Ammeter Reading, constant voltage) |
|---|---|---|
| Length is Doubled | Resistance doubles | Current decreases to one half |
| Length is Halved | Resistance is halved | Current increases two times |
| Cross-sectional Area is Doubled | Resistance is halved | Current increases two times |
| Cross-sectional Area is Halved | Resistance doubles | Current decreases to one half |
Beyond length and cross-sectional area, the resistance of a wire also depends on the material it's made of (resistivity) and its temperature. Different materials have different resistivities; for example, copper has a low resistivity, making it a good conductor. Temperature can also affect resistance; for most conductors, resistance increases with increasing temperature. All these factors collectively influence the current flowing through the wire for a given voltage, and thus impact the ammeter reading.
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