In an electric circuit, a wire of resistance 10 Ω is used. If this wire is stretched to a length double of its original value, the current in the circuit would become :
half of its original value.
double of its original value.
one-fourth of its original value.
four times of its original value.
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-1 – 2023
– Volume of the wire V = A × L. When stretched, the volume remains constant: $A’L’ = AL$.
– Ohm’s Law: Current I = Voltage V / Resistance R.
The wire is stretched to a new length $L’ = 2L$.
Since the volume remains constant, $A’L’ = AL$.
$A'(2L) = AL \Rightarrow A’ = \frac{A}{2}$.
The new resistance is $R’ = \rho \frac{L’}{A’} = \rho \frac{2L}{A/2} = \rho \frac{2L \times 2}{A} = 4 \rho \frac{L}{A}$.
So, $R’ = 4R$. The resistance becomes four times the original resistance.
Assuming the voltage V across the circuit element (or the voltage source) is constant:
Original current $I = \frac{V}{R}$.
New current $I’ = \frac{V}{R’} = \frac{V}{4R} = \frac{1}{4} \left(\frac{V}{R}\right) = \frac{1}{4}I$.
The current in the circuit would become one-fourth of its original value.