Four identical cells, each of emf 2 V and internal resistance 2 Ω, are joined in parallel. This combination is connected across a 1.5 Ω resistor. The current in the resistor is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 A
Cells joined in parallel act as one equivalent cell. For identical cells the emf stays the same, but the internal resistance becomes much smaller.
\( \begin{aligned} \frac{1}{r_{eq}} &= \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2} \\ &= 2\ \Omega^{-1} \\ r_{eq} &= 0.5\ \Omega \end{aligned} \)
\( \begin{aligned} \varepsilon_{eq} &= r_{eq}\left(\frac{2}{2}+\frac{2}{2}+\frac{2}{2}+\frac{2}{2}\right) \\ &= 0.5\times4 \\ &= 2\ \mathrm{V} \end{aligned} \)
The equivalent cell now sends current through the external resistor.
\( \begin{aligned} I &= \frac{\varepsilon_{eq}}{R+r_{eq}} \\ &= \frac{2}{1.5+0.5} \\ &= 1\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Cells in Series and in Parallel – “If there are n cells of emf ε1, . . . εn and of internal resistances r1,... rn respectively, connected in parallel, the … combination is equivalent to a single cell of emf εeq and … internal resistance req, such that”
\( \frac{1}{r_{eq}}=\frac{1}{r_1}+\dots+\frac{1}{r_n} \)\( \qquad (3.58) \)
\( \frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\dots+\frac{\varepsilon_n}{r_n} \)\( \qquad (3.59) \)