Question

14. Two wires of the same metal, have the same area of cross section but their lengths in the
ratio of 3 : 1. What should be the ratio of current flowing through them respectively, when
the same potential difference is applied across each of their length?​

Answer 1

Step-by-step explanation:

Ratio of cross-sectional areas of the wires = 3 : 1 and resistance of thick wire (R1) = 10 Ω.

RESISTANCE =>

$p \times \frac{l}{a} \alpha \frac{1}{a}$

Therefore ,

$\frac{r1}{r2} = \frac{a1}{a2} = \frac{1}{3}$

or ,

R1=3R1=3*10=30Ω

and equivalent resistance of these two resistances in series combination = R1 + R2 = 30+ 10 = 40 Ω.

Answer 2

Answer:

ratio of current flowing is 1:3

Step-by-step explanation:

v=ir

v=voltage, i= current and r= resistance

thus I∝$\frac{1}{R}$

Resisntance ∝ Length of wire , R∝L

since I∝$\frac{1}{R}$ and R∝Length

thus I∝$\frac{1}{Length}$

let k be the constant, I=$\frac{K}{L}$

Ratio of length of  wire1 and wire2 i.e L:L' =3:1

Current in wire 1 ($I_{1}$)= $\frac{K}{L}$

Current in wire 2 ($I_{2}$)= $\frac{K}{L'}$

thus $\frac{I_{1} }{I_{2} }$= $\frac{K}{L}/ \frac{K}{L'} =\frac{L'}{L}$=$\frac{1}{3}$

therefore $I_{1} :I_{2} =1:3$