To determine the percentage drop in heat output, we first need to understand that the power output
P of an electrical device is proportional to the square of the voltage
V. Mathematically, this relationship is represented as:
P∝V2 or P= where
R is the resistance of the heating unit. In this problem, we need to compare the power output at two different voltages, 400 V and 160 V .
Let's denote the initial power output at 400 V as
P1 and the power output at 160 V as
P2. Given that the initial power output is 500 W , we can write:
P1=500WSince the power is proportional to the square of the voltage:
P1=k⋅V12and
P2=k⋅V22where
k is a proportionality constant. We can divide these equations to find the ratio of
P2 to
P1 :
=Substituting the given voltage values:
====0.16 This means that the power output at 160 V is 0.16 times the power output at 400 V . To find
P2, we multiply
P1 by 0.16 :
P2=0.16⋅500W=80W The percentage drop in heat output is then given by:
Percentage Drop
=()×100Substituting the values:
Percentage Drop
=()×100=()×100=84%Thus, the percentage drop in heat output is
84%. The correct answer is:
Option C -
84%.