Respuesta :

Given :

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.

To Find :

The current in the circuit.

Solution :

We know, resistance of the ammeter is in series with the circuit.

So, total resistance is :

R = 2.5 + 0.5 ohm

R = 3 ohm

Also, e.m.f applied is 1.5 V .

Now, by ohm's law :

[tex]I = \dfrac{V}{R}\\\\I = \dfrac{1.5}{3}\\\\I = 0.5 \ A[/tex]

Therefore, the current in the circuit is 0.5 A.