Question

When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

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, $r = 1 \Omega$