Current Electricity: Real-World Scenarios (6)

Question

A household geyser is rated $2000$ W at $220$ V. Due to a voltage fluctuation in a Mumbai apartment, the supply drops to $200$ V. Find (a) the resistance of the heating element (assumed constant), (b) the actual power consumed at 200 V, and (c) the percentage drop in the heating rate. This is a JEE Advanced 2023 style question.

Solution — Step by Step

Why This Works

For a fixed-resistance device like a geyser or bulb, power scales as $V^2$. A small drop in voltage causes a much larger drop in heating rate. A 9% voltage drop ($220 \to 200$) gives a 17% power drop — almost double the percentage.

This is why your geyser feels slow during peak hours when voltage sags, and why bulbs visibly dim. The $V^2$ dependence amplifies every voltage variation.

Alternative Method

Use the ratio shortcut directly:

$\frac{P'}{P} = \left(\frac{V'}{V}\right)^2 = \left(\frac{200}{220}\right)^2 = \left(\frac{10}{11}\right)^2 = \frac{100}{121}$

$P' = 2000 \times \frac{100}{121} \approx 1653 \text{ W}$

This skips computing $R$ entirely — useful in MCQs where time matters.