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A wire when connected to 220 V mains sup...

A wire when connected to 220 V mains supply has power dissipation `P_1`. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is `P_2`. Then `P_2: P_1` is

A

1

B

4

C

2

D

3

Text Solution

Verified by Experts

Case I
Using the formula `P=V^2/R`
where, R is resistance of wire, V is voltage across wire and P is power dissipation in wire and `R=(pl)/A`
From equations (i) and (ii) , we have `p_1=V^2/(pl//A) or p_1=V^2/(pl)A`
Case II
Let `R_2` be net resistance of two wires in parallel, then `R_2=(R times R)/(R+R)=R/`
where R, is the resistance of half wire
`implies R_2=(p.(1/2))/(A.2) =(pl)/(4A) or p_2=V^2/(pl) 4A`.....(iv)
Hence from equations (iii) and (iv) we get `P_1/P_2=1/4 implies P_2/P_1=4/1`
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