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 Ω. The total resistance of the combination is

Question

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 Ω. The total resistance of the combination is (A) 5 / 2 Ω (B) 40 / 3 Ω (C) 40 Ω (D) 100 Ω

Tardigrade Admin · Accepted Answer

For the same length and same material, (R2/R1)=(A1/A2)=(3/1) or R2=3 R1 The resistance of thick wire, R1=10 Ω The resistance of thin wire =3 R1=3 × 10=30 Ω Total resistance =10+30=40 Ω

Q. 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 $10 Ω$ . The total resistance of the combination is

$5/2 Ω$

$40/3 Ω$

$40 Ω$

$100 Ω$

For the same length and same material,
$\frac{R _{2}}{R _{1}} = \frac{A _{1}}{A _{2}} = \frac{3}{1}$ or $R_{2} = 3 R_{1}$
The resistance of thick wire, $R_{1} = 10 Ω$
The resistance of thin wire $= 3 R_{1} = 3 \times 10 = 30 Ω$
Total resistance $= 10 + 30 = 40 Ω$

Q. 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 10Ω. The total resistance of the combination is

5/2Ω

40/3Ω

40Ω

100Ω