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.
