Question

If the voltage supply to an appliance is reduced to half of its voltage rating, what would be the effect on the power consumed by it?

Answer 1

Step 1: Given data

Power is directly proportional to the square of the voltage and inversely proportional to resistance.

Here resistance would be the same for both cases which is $R$.

Step 2: Formula used when voltage is V₁

Power ($P_1$) of the appliance when voltage $V_1$ is supplied

$P_1=\frac{{\left(V_1\right)}^2}{R}$

Step 3: Formula used when voltage is V₂

Power ($P_2$) of the appliance when voltage $V_2$ is supplied

$P_2=\frac{{\left(V_2\right)}^2}{R}$

Step 3: Calculation

Calculating $\frac{P_1}{P_2}$, on putting all the values we have,

$\frac{P_1}{P_2}=\frac{{\displaystyle \frac{{\left(V_1\right)}^2}{R}}}{{\displaystyle \frac{{\left(V_2\right)}^2}{R}}}={\left(\frac{V_1}{V_2}\right)}^2$

According to the question, $V_2=\frac{V_1}{2}$

Thus, $\frac{V_2}{V_1}=\frac{1}{2}$

Now calculating $\frac{P_2}{P_1}$, we will have

$\frac{P_2}{P_1}={\left(\frac{V_2}{V_1}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

Hence, If the voltage supply to an appliance is reduced to half of its voltage rating, then the power consumed by it will become one-fourth.