A cell supplies a current of 0.6 A through a \( 3\,\Omega \) resistor and a current of 0.2 A through a \( 10\,\Omega \) resistor. The internal resistance of the cell is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.5\,\Omega \)
For a cell of emf \( E \) and internal resistance \( r \) the current through an external resistance \( R \) is \( I = E/(R + r) \), so \( E = I(R + r) \). The emf is the same in both cases, so equate the two.
\( \displaystyle 0.6\,(3 + r) = 0.2\,(10 + r) \)
\( \displaystyle 1.8 + 0.6\,r = 2 + 0.2\,r \)
\( \displaystyle 0.4\,r = 0.2 \;\Rightarrow\; r = 0.5\,\Omega \)