Question

When a daniel cell of emf 1.08 volt is connected to an external resistor of 2 ohms, the potential difference falls to 0.9 volt. Calculate the internal resistance of the cell *​

Answer 1

Answer:

⇒E−Ir=0.6

Ir=0.48

E=I(R+r)

3+r

1.08

=I

⇒

3+r

1.08r

=0.48

1.08r=0.48r+0.48×3

0.6r=0.48×3

r=

6

4.8×3

r=2.4Ω

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