Two wires of same metal have the same length but their cross- sections are in the ratio 3 :1 . They are joined in series. The resistance of the thicker wire is `10 Omega` . The total resistance of the combination will be

Options

A. `(5//2)Omega`
B. `(40//3)Omega`
C. `40Omega`
D. `100Omega`

Answer

C

Explanation

For the same length and same material,
`(R_(2))/(R_(1))=(A_(1))/(A_(2))=(3)/(1)orR_(2)=3R_(1)`
The resistance of thicker wire, `R_(1)=10Omega`
The resistance of thinner wire = `3R_(1)=3xx10=30Omega`
Total resistance = `10Omega+30Omega=40Omega`