Q.Two cells of EMF $\varepsilon_1 = 1.5\ \text{V}$ (internal resistance $r_1 = 0.5\ \Omega$) and $\varepsilon_2 = 2.0\ \text{V}$ (internal resistance $r_2 = 0.3\ \Omega$) are joined in series, aiding each other, and the combination is connected to an external resistor $R = 4.2\ \Omega$. Find the current in the circuit.

The equivalent emf of parallel cells is the conductance-weighted average: E_eq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂). Plugging in gives 2·6 V. Option (d).

Step 1 — parallel-cell formula (from Kirchhoff's rules, NCERT/CBSE Class 12 Physics, Current Electricity):

E_eq = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂).

Step 2 — substitute E₁ = 3 V, r₁ = 0·2 Ω; E₂ = 2 V, r₂ = 0·3 Ω: …

For a mixed (series–parallel) grouping of cells, current is maximum when the external resistance equals the bank's internal resistance nr/m. Here nr/m = 6r/3 = 2r, so R : r = 2 : 1. Option (d).

Step 1 — arrangement: n = 6 cells in series per row, m = 3 rows in parallel (6 × 3 = 18 cells).

Step 2 — internal resistance of the whole bank: each row has 6r; three such rows in parallel give (6r)/3 = 2r.

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