Two sources of equal emf are connected in series. This combination is connected to an external resistance R. The internal resistance of two sources are \( r_1 \) and \( r_2 \) \( (r_2 \gt r_1) \). If the potential difference across the source of internal resistance \( r_2 \) is zero, then R equals to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( R = r_2 - r_1 \)
The two sources aid each other in series, so the current is the total emf divided by the total resistance:
\( \displaystyle I = \frac{2E}{R + r_1 + r_2} \)
The terminal potential difference of the second source is \( E - I r_2 \); it is zero when
\( \displaystyle E = I r_2 = \frac{2E\,r_2}{R + r_1 + r_2} \)
\( \displaystyle R + r_1 + r_2 = 2r_2 \;\Rightarrow\; R = r_2 - r_1 \)
Since \( r_2 \gt r_1 \), this resistance is positive, as it must be.
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
Current Electricity – “… 3.25 Two cells of same emf E but internal resistance r1 and r2 are connected in series to an external resistor R …. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.”