Question

When a resistance of $2\Omega$ is connected across a cell a current of $0.5$amp flows. If the resistance is increased to $5\Omega$ the current increase to $0.25$ ampere. The internal resistance of the cell is:

• A. $0.5\Omega$
• B. $1.0\Omega$
• C. $1.5\Omega$
• D. $2.0\Omega$

Answer 1

Answer: (B)

$1.0\Omega$

When a resistance of 2Ω is connected across the terminals of a cell, the current is 0.5 A. When resistance is increased to 5Ω, the current is 0.25 A.

Let the e.m.f. of cell be $E$ and internal resistance be $r$. Then

$0.5=\frac{E}{(r+2)}$ and $0.25=\frac{E}{(r+5)}$

On dividing, $2=\frac{5+r}{2+r}\Rightarrow r=1\Omega$