Question

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.

Answer 1

Answer:

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;

$The \ current \ in \ the \ circuit, \ I = \dfrac{E}{R + r}$

Substituting the known values, gives;

$I = \dfrac{1.5 \ V}{7 \ \Omega + 0.5 \ \Omega + 2.5 \ \Omega} = \dfrac{1.5 \ V}{10 \ \Omega} = 0.15 \ A$

The current in the circuit, I = 0.15 Ampere.