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. Calculat...

Assessment: WAEC SSCE - Physics - 1990 (Objective) Subject: Physics

Question 1 Report

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.
Answer Details
To calculate the current in the circuit, we need to use Ohm's law and Kirchhoff's circuit laws. First, we can calculate the total resistance of the circuit by adding the resistance of the ammeter and the resistor in series: R = 0.59 Ω + 7.09 Ω = 7.68 Ω Using Kirchhoff's circuit laws, we know that the current I through the circuit is equal to the e.m.f. E of the cell divided by the total resistance R of the circuit plus the internal resistance r of the cell: I = E / (R + r) Substituting the given values, we get: I = 1.5 V / (7.68 Ω + 2.59 Ω) = 0.15 A Therefore, the current in the circuit is 0.15 A. So, the answer is 0.15 A.

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