A battery of e.m.f 10V an internal resistance 2Ω is connected to a resistance of…

Physics WAEC 1994

A battery of e.m.f 10V an internal resistance 2Ω is connected to a resistance of 6Ω. Calculate the p.d

  • 8.75V
  • 7.50V checkmark
  • 5V
  • 2.50V
  • 1.25V

The correct answer is: B

Explanation

V = E - Ir = 10V - \(\frac{10 \times 2}{8}\) = 7.5V
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