Question

A cell of Emf 2V and internal resistance 1.2Ω is connected with an ammeter of resistance 0.8 Ω and two resistors of 4.5Ω and 9Ω as shown in the diagram below:

1. What would be the reading on the Ammeter? 

2. What is the potential difference across th . terminals of the cell?

Answer 1

Given that =2V, r = I.2Ω, RA = O.8Ω,R₁ = 4.5Ω , R₂=9Ω ]

1. We know that for the circuit = IR total 

Now, the total resistance of the circuit is 

R total = r + RA + R_p 

1/R_p = 1/4.5 + 1/9 = 3/9 

R_p = 3Ω 

R total = 1.2 + 0.8 + 3 = 5Ω 

Hence, the current through the ammeter is

I = V/R_total = 2/5 = 0.4A

(ii) I = (E - V)/r

⇒ 0.4 = (2 - V)/1.2

I = v/R total = 2/5 = 0.4 AB

⇒ 0.48 = 2 – V - I 

V = 2 – 4.8 – 1.52V 

:. Potential difference Vcell 1.52V