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ICSE Class 10 Physics 2015 Solved Paper

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A cell of Emf 2V and internal resistance 1.2Ω is connected with an ammeter of resistance 0.8Ω and two resistors of 4.5Ω and 9Ω as shown in the diagram below :
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Question : 59
Total: 74
What would be the reading on the Ammeter?
Solution:  
Given : E=2V,r=1.2Ω,R= (external resistance)
Let 4.5Ω and 9Ω connected in parallel, then equivalent resistance
‌‌
1
Rp
=‌
1
4⋅5
+‌
1
9
‌‌
1
Rp
=‌
2+1
9
=‌
3
9
=‌
1
3
‌Rp=3Ω
Now 0.8Ω and Rp resistance in series, then total resistance
R=3+0.8+1⋅2Ω=5Ω
Current in the ammeter
I‌=‌
‌ Total e.m.f. ‌
‌ Total resistance ‌
‌=‌
2
5
=0.4A
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