A cell of e.m.f. 1.5 V and internal resistance of 2.59 is connected in series…

Physics WAEC 1990

A cell of e.m.f. 1.5 V and internal resistance of 2.59 is connected in series with an ammeter of resistance 0.59 and a resistor of resistance 7.09. Calculate the current in the circuit.

  • 0.15 A checkmark
  • 0.20 A
  • 0.60A
  • 3.00A
  • 6.67 A.

The correct answer is: A

Explanation

I =
E/R + r
=
1.5/2.5 + 0.5 + 7
= 0.15
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