Seven lead accumulators, each of emf \( 2\,\mathrm{V} \), and internal resistance \( 0.07\,\Omega \), are connected in series with an external resistance \( 9.51\,\Omega \). The current in the circuit is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.4\,\mathrm{A} \)
When cells are joined in series, their emfs add and so do their internal resistances. The current then follows from Ohm's law for the whole circuit.
\( \displaystyle \varepsilon_{\mathrm{net}} = 7 \times 2 = 14\,\mathrm{V} \)
\( \displaystyle r_{\mathrm{net}} = 7 \times 0.07 = 0.49\,\Omega \)
\( \displaystyle I = \frac{\varepsilon_{\mathrm{net}}}{R + r_{\mathrm{net}}} = \frac{14}{9.51 + 0.49} = \frac{14}{10} = 1.4\,\mathrm{A} \)