Two batteries of emf 5 V with internal resistance \( 2\,\Omega \) and 6 V with internal resistance \( 3\,\Omega \) are connected in series with an external resistor. If the current in the circuit is 0.5 A, then the resistance of the resistor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 17\,\Omega \)
Cells joined in series (positive to negative) add their emfs and their internal resistances. This is the standard reading; if the cells opposed each other the net emf would be \( 1\,\mathrm{V} \), and a current of \( 0.5\,\mathrm{A} \) would need a total resistance of only \( 2\,\Omega \), less than the \( 5\,\Omega \) of the cells, which is impossible.
\( \varepsilon_{\mathrm{eq}} = 5 + 6 = 11\,\mathrm{V} \)
\( r_{\mathrm{eq}} = 2 + 3 = 5\,\Omega \)
The current in the circuit with external resistance \( R \) is
\( \displaystyle I = \frac{\varepsilon_{\mathrm{eq}}}{R + r_{\mathrm{eq}}} \;\Rightarrow\; R = \frac{\varepsilon_{\mathrm{eq}}}{I} - r_{\mathrm{eq}} \)
\( \displaystyle R = \frac{11}{0.5} - 5 = 22 - 5 = 17\,\Omega \)