Question

Three wires of same material are connected in parallel to a source of emf. The length ratio of the wires is 1 : 2 : 3 and the ratio of their area of cross section is 2 : 4 : 1.

Table 1	Table 2
(a)Resistance ratio	(p) 6 : 6 : 1
(b) Current ratio	(q) 1 : 6 : 6
(c) Power ratio	(r) 1 : 1 : 6
	(s) None

• A. 1) a-r,b-q,c-p
• B. 2) a-p,b-q,c-r
• C. 3) a - r; b - p, c - p
• D. 4) a-q,b-p,c-r

Answer 1

The correct option is C 3) a - r; b - p, c - p

$\Rightarrow R_1:R_2:R_3=\frac{L_1}{A_1}:\frac{L_2}{A_2}:\frac{L_3}{A_3}$

$=\frac{1}{A_1}:\frac{L_2}{2A_1}:\frac{L_3}{\frac{A_1}{2}}$

$=1:\frac{2}{2}:\frac{3}{\frac{1}{2}}$

$R_1:R_2:R_3=1:1:6$

As they are in parallel $I\alpha \frac{1}{R}$

$\Rightarrow I_1:I_2:I_3=\frac{1}{R_1}:\frac{1}{R_2}:\frac{1}{R_3}$

$=1:1:\frac{1}{6}=6:6:1$

$Power=VI\Rightarrow P\alpha I$

$\Rightarrow P_1:P_2:P_3=I_1:I_2:I_3$

$=6:6:1$