A wire of resistance 0.25 ohm is first connected to a cell of internal...
Calculating Current with Single Cell:
When the wire is connected to a single cell with an internal resistance, the total resistance in the circuit would be the sum of the wire resistance and the internal resistance of the cell. Therefore, the total resistance would be 0.25 ohm + 0.50 ohm = 0.75 ohm.
Using Ohm's Law (V = IR), we can find the current (I) in the circuit. Since the cell provides the voltage, let's assume it is 1V for simplicity. Therefore, I = V / R = 1V / 0.75 ohm = 1.33 A.
Calculating Current with 10 Cells in Parallel:
When the wire is connected to 10 cells in parallel, the total resistance in the circuit will decrease due to the parallel combination. The resistance of each cell remains 0.50 ohm, but when connected in parallel, the total resistance becomes 0.50 ohm / 10 = 0.05 ohm.
Now, the total resistance in the circuit will be the sum of the wire resistance and the total resistance of the parallel cells, which is 0.25 ohm + 0.05 ohm = 0.30 ohm.
Using Ohm's Law again, we can find the current (I) in the circuit. With the same voltage of 1V, I = V / R = 1V / 0.30 ohm = 3.33 A.
Ratio of Currents:
The ratio of currents in the two cases can be calculated by dividing the current with 10 cells in parallel by the current with a single cell. Therefore, the ratio of currents is 3.33 A / 1.33 A = 2.5, which can be simplified to 5:2.
Therefore, the correct answer is 4) 2:5.
A wire of resistance 0.25 ohm is first connected to a cell of internal...
4) 2:5
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