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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 9.18 V. When only the car's lights are used, the current through the battery is 2.10 A and the terminal potential difference is 12.6 V. Find the battery's emf.

Respuesta :

lublana

Answer:

12.74 V

Explanation:

We are given that

Current, I1=54 A

Potential difference, V1=9.18V

I2=2.10 A

V2=12.6 V

We have to find the battery's emf.

[tex]E=V+Ir[/tex]

Using the formula

[tex]E=9.18+54r[/tex] ....(1)

[tex]E=12.6+2.10r[/tex]  .....(2)

Subtract equation (1) from (2)

[tex]0=3.42-51.9r[/tex]

[tex]3.42=51.9r[/tex]

[tex]r=\frac{3.42}{51.9}=0.0659ohm[/tex]

Using the value of r in equation (1)

[tex]E=9.18+54(0.0659)[/tex]

[tex]E=12.74 V[/tex]