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ICSE Class 10 Physics 2020 Paper

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The diagram above shows three resistors connected across a cell of e.m.f. 1.8V1.8 V and internal resistance rr. Calculate:
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Question : 53 of 67
Marks: +1, -0
The internal resistance rr.
Solution:  
In the given diagram, 3 Ω3 \, \Omega and 1.5 Ω1.5 \, \Omega resistor are in parallel.
  ∴  1Rp  =  13+  11.5\;\therefore\;\frac{1}{R_p}\;=\;\frac{1}{3}+\;\frac{1}{1.5}
1Rp  =  1+23=  33\frac{1}{R_p}\;=\;\frac{1+2}{3}=\;\frac{3}{3}
  ∴  Rp  =1 Ω\;\therefore\;R_p\;=1 \, \Omega
Now RpR_p and 4 Ω4 \, \Omega resistances are in series
∴    Rs=1+4=5 Ω\therefore\;\;R_s=1+4=5 \, \Omega
So net external resistance of the circuit =5 Ω=5 \, \Omega
   We know,     I=ERext+r\;\text{ We know, }\;\;I=\frac{E}{R_{\text{ext}}+r}
  ⇒    0.3=1.85+r\;\Rightarrow\;\;0.3=\frac{1.8}{5+r}
  ⇒    0.3(5+r)=1.8\;\Rightarrow\;\;0.3(5+r)=1.8
  ⇒    (5+r)=6\;\Rightarrow\;\;(5+r)=6
  ⇒    r=6−5\;\Rightarrow\;\;r=6-5
  r=1 Ω\; r=1 \, \Omega
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