4 cell each of emf \(2 \mathrm{v}\) and internal resistance of \(1 \Omega\) are connected in parallel to a load resistor of \(2 \Omega\) Then the current through the load resistor is.... (A) \(2 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(1 \mathrm{~A}\) (D) \(0.888 \mathrm{~A}\)

When cells are connected in parallel, the emf remains the same, but the internal resistance decreases. For 'n' cells connected in parallel, the equivalent internal resistance (R_eq) is given by: \[R_{eq} = \frac{R}{n}\] In our case, there are 4 cells with internal resistance of 1 ohm each, so, \[R_{eq} = \frac{1}{4}\] Thus, the equivalent emf is 2V and the equivalent internal resistance is \(0.25\Omega\).

Now that we have the equivalent emf and internal resistance, we can treat the parallel combination of cells as a single cell with these values. The total resistance of the circuit will be the sum of the load resistor and the equivalent internal resistance. According to Ohm's law, the current (I) passing through the load resistor would be: \[I = \frac{E_{eq}}{R_{eq} + R_L}\] where \(E_{eq}\) is the equivalent emf (2V), \(R_{eq}\) is the equivalent internal resistance (\(0.25\Omega\)), and \(R_L\) is the load resistor (2 ohms): \[I = \frac{2}{0.25 + 2}\] Calculating this, we get: \[I = \frac{2}{2.25}\] \[I = 0.888 \mathrm{~A}\] Thus, the current through the load resistor is \(0.888 \mathrm{~A}\), which corresponds to option (D).