Six identical cells, each of emf 3 V and internal resistance of r, are connected in series to a \( 3\,\Omega \) resistance. If the net current flowing in the circuit is 2.0 A. The internal resistance of the cell is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1\,\Omega \)
For cells in series, the emfs add and the internal resistances add. The current then follows from Ohm's law for the whole circuit.
\( \varepsilon_{\mathrm{eq}} = 6 \times 3 = 18\,\mathrm{V} \)
\( r_{\mathrm{eq}} = 6r \)
\( \displaystyle I = \frac{\varepsilon_{\mathrm{eq}}}{R + r_{\mathrm{eq}}} \;\Rightarrow\; 2 = \frac{18}{3 + 6r} \)
\( \displaystyle 3 + 6r = 9 \;\Rightarrow\; r = 1\,\Omega \)