Question

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.

Answer 1

12.74 V

Step-by-step 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.

`E=V+Ir`

Using the formula

`E=9.18+54r` ....(1)

`E=12.6+2.10r` .....(2)

Subtract equation (1) from (2)

`0=3.42-51.9r`

`3.42=51.9r`

`r=(3.42)/(51.9)=0.0659ohm`

Using the value of r in equation (1)

`E=9.18+54(0.0659)`

`E=12.74 V`