Question

A \[500\,W\] heating unit is designed to operate on a \[115\,V\] line. If line voltage drops to 110 V line, the percentage drop in the heat output will be:
(A) $7.6%$
(B) $8.5%$
(C) $8.1%$
(D) $10.2%$

Answer 1

Hint: Using Joule’s law, express the heat energy output of the heating unit. Use Ohm’s law to rewrite the quantities in terms of voltage and resistance. The resistance of the heating unit will not change when there is voltage drop in the line. Therefore, equate the resistance in both cases.

Formula used:
Joule’s law,
\[\dfrac{Q}{t} = {I^2}R\]
Here, t is the time, I is the current and R is the resistance.

Complete step by step answer:
We have given the power rating of the heating unit and it is 500 W.
We know Joule’s law of heating states that the quantity of heat produced in the conductor per unit time is equal to the product of square of the current through the conductor and resistance of the conductor.
\[\dfrac{Q}{t} = {I^2}R\]
Here, t is the time, I is the current and R is the resistance.
We know that the flow of energy per unit time is power. Therefore, we can express the above equation as follows,
\[P = {I^2}R\]
From Ohm’s law, we have, \[I = \dfrac{V}{R}\]. Therefore, using this in the above equation, we get,
\[P = \dfrac{{{V^2}}}{R}\]
We can express the power for the line of voltage 115 V as follows,
\[{P_1} = \dfrac{{V_1^2}}{R}\] …… (1)
Also, the power for the line of voltage 110 V is,
\[{P_2} = \dfrac{{V_2^2}}{R}\] …… (2)
We have given that the line voltage is dropped in the same line. Even if the voltage drops, the resistance of the unit will remain constant. Therefore, we can write,
\[R = \dfrac{{V_1^2}}{{{P_1}}} = \dfrac{{V_2^2}}{{{P_2}}}\]
\[ \Rightarrow {P_2} = {P_1}\dfrac{{V_2^2}}{{V_1^2}}\]
We substitute 115 V for \[{V_1}\], 110 V for \[{V_2}\] and 500 W for \[{P_1}\] in the above equation.
\[{P_2} = \left( {500} \right)\dfrac{{{{\left( {110} \right)}^2}}}{{{{\left( {115} \right)}^2}}}\]
\[ \Rightarrow {P_2} = \left( {500} \right)\left( {0.915} \right)\]
\[ \Rightarrow {P_2} = 457.46\,W\]
Now, the percentage change in the heat output is,
\[\left( {\dfrac{{500 - 457.46}}{{500}} \times 100} \right) = 8.5\% \]
Therefore, the output power will decrease by 8.5% and therefore, the heating output will also drop by 8.5%.

So, the correct answer is option (B).

Note:Students can use Ohm’s law wherever there is term voltage and current resistance to relate them. The power rating and the heat output is the same for the heating unit. To calculate the percent change, it should be the ratio of change in the heat energy to the initial heat energy.