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

Options

A. `10 Omega`
B. `20 Omega`
C. `40 Omega`
D. `100 Omega`

Answer

C

Explanation

Length of each wire = l , Area of thick wire `(A_1)= 3 A`, Area of thin wire `(A_2)= A` and resistance of thick wire `(R_1) = 10 Omega`. Resistance `(R) = rho (l)/( A) prop (1)/( A)` (if `l` is constant)
`therefore (R_1)/(R_2) = (A_2)/(A_1) = (A)/( 3 A) = (1)/(3)`
or, `R_2= 3 R_1 = 3 xx 10= 30 Omega`
The equivalent resistance of these two resistors in series
`= R_1 + R_2 = 30 + 10 = 40 Omega`.