ICSE Class 10 Physics 2015 Solved Paper

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

Question : 59 of 74

Marks: +1, -0

What would be the reading on the Ammeter?

Solution:

Given :

{\E}=2 {\V}, r=1.2 Ω, {\R}=

(external resistance)
Let

4.5 Ω

and

9 Ω

connected in parallel, then equivalent resistance

\;\;{1}/{ {\R}_p}=\;{1}/{4 ⋅ 5}+\;{1}/{9}

\;\;{1}/{ {\R}_p}=\;{2+1}/{9}=\;{3}/{9}=\;{1}/{3}

\; {\R}_p=3 Ω

Now

0.8 Ω

and

{\R}_p

resistance in series, then total resistance

{\R}=3+0.8+1 ⋅ 2 Ω=5 Ω

Current in the ammeter

I\;=\;{\;\text " Total e.m.f. "\;}/{\;\text " Total resistance "\;}

\;=\;{2}/{5}=0.4 {\A}

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