A cell of e.m.f. 1.5V and internal resistance of 2.5Ω is connected in series with…

Physics WAEC 1994

A cell of e.m.f. 1.5V and internal resistance of 2.5Ω is connected in series with an ammeter of resistance 0.5W and a resistor of resistance of 7.0Ω. Calculate the current in the circuit.

  • 0.15A checkmark
  • 0.20A
  • 2.10A
  • 3A
  • 6.6A

The correct answer is: A

Explanation

I =\(\frac{E}{R+r} = \frac{1.5}{2.5+0.5+7}\) = 0.15A
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