Question
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
Answer :
4
Solution :
Case 1: $${P_1} = \frac{{{V^2}}}{R}$$
Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $$\frac{R}{2}.$$ These are connected in parallel
$$\eqalign{ & \therefore {R_{eq}} = \frac{{\frac{R}{2}}}{2} = \frac{R}{4} \cr & \therefore {P_2} = \frac{{{V^2}}}{{\frac{R}{4}}} = 4\left( {\frac{{{V^2}}}{R}} \right) = 4{P_1} \cr} $$
Case 1: $${P_1} = \frac{{{V^2}}}{R}$$
Case 2 : The wire is cut into two equal pieces. Therefore the resistance of the individual wire is $$\frac{R}{2}.$$ These are connected in parallel
$$\eqalign{ & \therefore {R_{eq}} = \frac{{\frac{R}{2}}}{2} = \frac{R}{4} \cr & \therefore {P_2} = \frac{{{V^2}}}{{\frac{R}{4}}} = 4\left( {\frac{{{V^2}}}{R}} \right) = 4{P_1} \cr} $$
