Three similar ideal cells (no internal resistances) supply current of 0.5 Ampere to a 10 Ohms resistor when they are in the parallel.How much current will they supply to the same risistor when they are connected in series aiding?
To determine the current when the three similar ideal cells are connected in series aiding, let's break down the solution step by step:
1. First, we need to understand the concept of series aiding for batteries. When batteries are connected in series aiding, their voltages add up while the current remains the same.
- The voltage across each battery (V1, V2, and V3) is the same, and it will add up when connected in series.
- The current (I) remains constant.
2. Given that the three ideal cells are supplying a current of 0.5 Amperes when connected in parallel, we can assume that each cell supplies 0.5 Amperes.
- So, the current (I) supplied by each cell is 0.5 Amperes.
3. Now, let's calculate the voltage across each cell to determine the total voltage when connected in series aiding.
- The voltage (V) across each cell can be found using Ohm's law: V = I * R, where R is the resistance.
For each cell:
V1 = I * R = 0.5 Amps * 10 Ohms = 5 Volts
V2 = I * R = 0.5 Amps * 10 Ohms = 5 Volts
V3 = I * R = 0.5 Amps * 10 Ohms = 5 Volts
4. Since the cells are connected in series, the voltages add up.
Total Voltage (V_total) = V1 + V2 + V3 = 5 Volts + 5 Volts + 5 Volts = 15 Volts
5. Since the voltages add up and the resistance remains the same (10 Ohms), we can use Ohm's law again to calculate the current (I_total) supplied when the cells are connected in series aiding.
I_total = V_total / R
I_total = 15 Volts / 10 Ohms = 1.5 Amperes
Therefore, when the three ideal cells are connected in series aiding, they will supply a current of 1.5 Amperes to the 10 Ohms resistor.