Two identical cells of emf 2 V each joined in parallel provide supply to an external circuit containing two resistors of \( 20\,\Omega \) each joined in parallel. A very high resistance voltmeter reads the terminal voltage of the cells to be 1.6 V. The internal resistance of each cell is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5\,\Omega \)
Two identical cells in parallel behave as one cell of the same emf E = 2 V and internal resistance \( r/2 \), where r is the internal resistance of each. The terminal voltage of a cell delivering current I is \( V = E - Ir_{\mathrm{eq}} \).
External circuit
\( \displaystyle R = \frac{20\times20}{20+20} = 10\,\Omega \)
\( \displaystyle I = \frac{V}{R} = \frac{1.6\,\mathrm{V}}{10\,\Omega} = 0.16\,\mathrm{A} \)
Internal resistance
\( \displaystyle V = E - I\,\frac{r}{2} \)
\( \displaystyle 1.6 = 2 - 0.16\times\frac{r}{2} \Rightarrow 0.08\,r = 0.4 \)
\( r = 5\,\Omega \)
The cells are identical, so each supplies 0.08 A and drops \( 0.08\times5 = 0.4\,\mathrm{V} \) inside itself, which checks with \( 2 - 0.4 = 1.6\,\mathrm{V} \).