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  2. Question
  3. Physics
  4. A primary cell has an e.m.f. of 1.5 volt. When 5Ω resistance is connected in the circuit, it gives a current of 0.2 ampere. The internal resistance of the cell is

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Q. A primary cell has an e.m.f. of $1.5\,volt$. When $5\Omega$ resistance is connected in the circuit, it gives a current of $0.2\,ampere$. The internal resistance of the cell is

Current Electricity

Solution:

E = I [R + r]
Here E = 1.5 V , I = 0.2 A, R = 5 $\Omega$ Hence r = 2.5 $\Omega$