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?

E=I*R = 0.5 * 10=5 Volts, each battery.

I = E/R = 15/10 = 1.5 Amps.

Well, if we're talking about ideal cells, then we can assume they have imaginary personalities too. Let's picture them as celebrity chefs in a cooking show. When they're in parallel, they work as a team, like the Three Musketeers of current supply.

So, they each supply a current of 0.5 Ampere, totaling 1.5 Amperes in total, because they don't like to hog the spotlight individually. They share the fame and glory equally.

Now, in series aiding, it's like they're competing for the title of "Master Current Supplier." And just like in a cooking competition, they have to step up their game.

So, when they're connected in series, they combine their powers, and each ideal cell supplies the same current. Since there are three of them, they'll still supply a current of 0.5 Ampere each. But remember, they're working together, not against each other, so their currents add up.

Therefore, the total current supplied to the resistor when they're connected in series aiding will be 0.5 Ampere + 0.5 Ampere + 0.5 Ampere, which adds up to a total of 1.5 Amperes.

So, whether they're in parallel or series aiding, these ideal cells know how to deliver the goods and supply a current that's definitely shockingly good!

When the three ideal cells are connected in parallel, the total current supplied to the 10 Ohms resistor is 0.5 Ampere.

Since the cells are ideal, the voltage across the resistor remains the same in both parallel and series configurations.

In a parallel circuit, the total current is the sum of individual currents. Therefore, each cell supplies 0.5 Ampere / 3 = 0.1667 Ampere.

Now, when the three cells are connected in series aiding, the total current supplied to the 10 Ohms resistor is also 0.1667 Ampere.

Hence, the three cells, when connected in series aiding, will supply a current of 0.1667 Ampere to the same resistor.

To find the current supplied by the three similar ideal cells when connected in series aiding, we need to first determine the equivalent resistance of the circuit and then apply Ohm's Law formula.

When the cells are connected in parallel, the equivalent resistance is given by the reciprocal of the sum of the reciprocals of each individual resistance:

1/Req = 1/R1 + 1/R2 + 1/R3

Given that the resistance of the resistor is 10 Ohms and each cell supplies a current of 0.5 Ampere, the equivalent resistance is:

1/Req = 1/10 + 1/10 + 1/10
1/Req = 3/10

Taking the reciprocal on both sides, we get:

Req = 10/3 Ohms

Now, when the cells are connected in series aiding, the total resistance is the sum of the individual resistances:

Rtotal = R1 + R2 + R3

Given that all cells are identical and have no internal resistance, their resistance values are the same. Therefore:

Rtotal = R1 + R2 + R3 = 3R

Substituting the value of Rtotal:

10/3 = 3R
R = 10/9 Ohms

Now, we can apply Ohm's Law to find the current supplied by the cells:

I = V/R

Given that the cells are ideal, they will supply the same voltage. Therefore, we can choose any one of the three cells to find the voltage. Let's choose the first cell:

V = I * R1

V = 0.5 * 10/9
V = 5/9 Volts

Now, we can find the current supplied by the three cells when connected in series aiding:

I = V/Rtotal

I = (5/9) / (10/9)
I = 1/2 Ampere

Hence, the three similar ideal cells supply a current of 0.5 Ampere to a 10 Ohms resistor when connected in parallel, and they will supply a current of 1/2 Ampere to the same resistor when connected in series aiding.