A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.

What is the battery's internal resistance?

0.46Ω

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P =

=> R =             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ =    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ =

R₁ =

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ =    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ =

R₂ =

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

= +       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

= +

=

Rₓ =

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I =

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r =

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

Where:

As you can see, we don't know the exactly value of the . However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

Solving for :

Now, the electric power is given by:

Where:

So:

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

Finally, now we have all the data, let's replace it into the internal resistance equation:

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going   so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me