Question

A metal rod of square cross-sectional area A having length l has current I flowing through it when a potential difference of V volt is applied across its ends (figure I). Now the rod is cut parallel to its length into two identical pieces and joined as shown in figure II. What potential difference must be maintained across the length of 2l. so that the current in the rod is still I?

Answer 1

From Ohm's law, we have

\[V = IR\]

\[ \Rightarrow V = I\rho\frac{l}{A} . . . . . (1)\]

When the rod is cut parallel, and rejoined by length, the length of the conductor becomes 2l, whereas the area decrease to

\[\frac{A}{2}\]. If the current remains the same the potential changes as

\[V' = I\rho\frac{2l}{\frac{A}{2}} = 4 \times I\rho\frac{l}{A} = 4V\]

The new potential applied across the metal rod will 4 times the original potential (V).