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ICSE Class 10 Physics 2014 Paper

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Question : 27 of 66
Marks: +1, -0
Find the equivalent resistance between points AA and BB.
Solution:  
   Let       1R1=13+13+13=33=1\;\text{ Let }\;\;\;\frac{1}{R_1}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1
∴    R1=1Ω\therefore\;\; R_1=1 \Omega
R2  =5ΩR_2\;=5 \Omega
  1R3  =  14+  16=  3+212=  512\;\frac{1}{R_3}\;=\;\frac{1}{4}+\;\frac{1}{6}=\;\frac{3+2}{12}=\;\frac{5}{12}
  R3=  125=2.4Ω\; R_3=\;\frac{12}{5}=2.4 \Omega
Equivalent resistance between AA and BB,
R  =R1+R2+R3R\;=R_1+R_2+R_3
  =1+5+2.4=8.4Ω.\;=1+5+2.4=8.4 \Omega .
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