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When a car's starter is in use, it draws a large current. The car's lights draw much less current. As a certain car is starting, the current through the battery is 54.0 A and the potential difference across the battery terminals is 9.63 V. When only the car's lights are used, the current through the battery is 2.06 A and the terminal potential difference is 12.8 V. Find the battery's emf.

Respuesta :

Answer:

ε = 12.9257 V

Explanation:

To do this we need to see what is happening in both cases, when the lights are not being used and when the lights are used.

We will call case 1, the part when the lights are not being used, and case 2 when they're used.

Case 1:

Here, we should write the expression that will help us to determine the emf of the battery, which is:

ε - I₁R = V₁   (1)

Where:

ε: Emf of the battery

I: current of the car

R: Resistance

V: Voltage

Applying this expression with the given data, we have the following:

ε - 54R = 9.63    (2)

For the moment, we can't do anything because we have two incognites. Let's see what happen when we write the expression for case 2.

Case 2:

Applying the same as case 1 but using the other data we have:

ε - I₂R = V₂

Replacing we have:

ε - 2.06R = 12.8    (3)

Now, we can use (2) and (3) and solve for ε calculating the resistance:

ε = 12.8 + 2.06R   (4)

ε = 9.63 + 54R     (5)

Equalling both:

12.8 + 2.06R = 9.63 + 54R

12.8 - 9.63 = 54R - 2.06R

3.17 = 51.94R

R = 3.17 / 51.94 = 0.061 Ω

With this value, we can calculate the emf, using equation (4) or (5):

ε = 12.8 + 2.06(0.061)

ε = 12.9257 V

Hope this helps

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