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

The current in the circuit is 0.15 Ampere

Explanation:

The given parameters of the cell are;

The electromotive force (e.m.f.) of the cell, E = 1.5 V

The resistance of the cell, r = 2.5 ohm

The resistance of the ammeter = 0.5 ohm

The resistance of the resistor = 7.0 ohm

The formula for the e.m.f., E of a cell is given as follows;

e.m.f. E = I·(R + r)

Where;

I = The current in the circuit

R = The sum of the resistances in the circuit = 7.0 Ω + 0.5 Ω + 2.5 Ω = 10 Ω

Therefore, we have;

Substituting the known values, gives;

The current in the circuit, I = 0.15 Ampere.

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