Case: Read the passage and answer the question.
A cell of unknown emf E and internal resistance r, two unknown resistances \( R_1 \) and \( R_2 \) \( (R_2\gt R_1) \) and a perfect ammeter are given. The current in the circuit is measured in five different situations:
(i) without any external resistance in the circuit,
(ii) with resistance \( R_1 \) only,
(iii) with resistance \( R_2 \) only,
(iv) with both \( R_1 \) and \( R_2 \) used in series combination, and
(v) with both \( R_1 \) and \( R_2 \) used in parallel combination.
The currents obtained in the five cases are 0.42 A, 0.6 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order.
In terms of r, what is the emf of the cell?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.2 r
For a cell of emf \( \varepsilon \) and internal resistance \( r \) connected to an external resistance \( R \), the current is \( I=\varepsilon/(R+r) \). Any external resistance, whether \( R_1 \), \( R_2 \), or their series or parallel combination, adds to \( r \) and lowers the current. So the largest current is drawn when there is no external resistance \( (R=0) \).
Of the five currents, the largest is 4.2 A, so it belongs to situation (i):
\( \displaystyle I_{\max}=\frac{\varepsilon}{r} \)
\( \begin{aligned} \varepsilon &= I_{\max}\,r \\ &= 4.2\,r \end{aligned} \)
Here \( \varepsilon \) is in volts when \( r \) is in ohms.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “…
\( \displaystyle I=\frac{\varepsilon}{R+r} \qquad (3.40) \)
The maximum current that can be drawn from a cell is for R = 0 and it is Imax = ε/r.”