Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1 . They are joined in series. The resistance of thicker wir
Two wires of the same metal have same length, but their cross-sections are in the ratio 3:1 . They are joined in series. The resistance of thicker wire is `10 Omega`. The total resistance of the combination will be
A. `2.5Omega`
B. `40//3Omega`
C. `40Omega`
D. `100Omega`
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Correct Answer - C
`(R_(1))/(R_(2))=((rho_(1) l_(1)//A_(1)))/((rho_(2)l_(2)//A_(2)))=(A_(2))/(A_(1))`
`:.R_(2) =(A_(1))/(A_(2))xxR_(1)=(3xx10)/(1)= 30 Omega`
Total resistance of combination `=R_(1)+R_(2)=40 Omega`
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