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 62.4 A and the potential difference across the battery terminals is 9.90 V. When only the car's lights are used, the current through the battery is 2.18 A and the terminal potential difference is 11.1 V.

Find the battery's emf.


Find the internal resistance.

Answer 1

Solution)

We know,

E=V+Ir

E =9.9+62.4r

E-62.4r=9------1

Next, when car lights are on

E=11.1+2.18r

E-2.18r=11.1----2

Now, equation 1- equation 2

E-62.4r-E+2.18r =9-11.1

60.22 r=2.1

r=0.0348 ohms (Ans)

Substituting r in eq 1

E-62.4*0.0348=9

E=11.17 V (Ans)

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