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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 10Ω. The total resistance of the combination will be
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- 10 Ω
- 20 Ω
- 40 Ω
- 100 Ω
Correct Option: C
Length of each wire = l; Area of thick wire (A1) = 3A; Area of thin wire (A2) = A and resistance of thick wire (R1) = 10 Ω.
| Resistance (R) = ρ | ∝ | (if l is constant) | ||
| A | A |
| ∴ | R1 | = | A2 | = | A | = | 1 | ||
| R2 | A1 | 3A | 3 |
or, R2 = 3R1 = 3 × 10 = 30 Ω
The equivalent resistance of these two resistors in series
= R1 + R2 = 30 + 10 = 40Ω.
