First slide

combination of resistance

Question

Two wires of 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 $Ω$ .The total resistance of the combination will be

Moderate

A

$\frac{5}{2} Ω$
B

$\frac{40}{3} Ω$
C

$40 Ω$
D

$100 Ω$

Solution

We know that R= p(l/A)
For same material and same length
$\frac{R_{1}}{R_{2}} = \frac{A_{2}}{A_{1}} = \frac{3}{1} or R_{1} = 3 {\dot{R}}_{2}$
Given, $R_{2} = 10 Ω, hence R_{1} = 30 Ω$
In series, R = R₁ +R₂
$= 30 Ω + 10 Ω = 40 Ω$

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