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.
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Correct Answer: Option A
Explanation:
I =\(\frac{E}{R+r} = \frac{1.5}{2.5+0.5+7}\) = 0.15A
I =\(\frac{E}{R+r} = \frac{1.5}{2.5+0.5+7}\) = 0.15A