Question

A lamp is marked 60 W, 220 V. If it operates at 200 V, the rate of consumption of energy will _____

• A. decrease
• B. increase
• C. first increase then decrease
• D. remain unchanged

Answer 1

Answer: (A)

Given power of bulb =60W

Voltage=220V

Power $=\frac{V^2}{R}$

$60=\frac{220\times 220}{R}$

$R=\left(\frac{220\times 220}{60}\right)\Omega$

$R=\left(\frac{220\times 22}{6}\right)\Omega$ (1)

When voltage = 200V, resistance will be the same

$P=\frac{200\times 200}{R}$

From 1,

$=\frac{200\times 200\times 6}{220\times 22}$

$\frac{200\times 20\times 6}{22\times 22}$

$=49.5W$

We see that power gets decreased.

Power means rate of consumption of energy, so rate of consumption of energy decreases.

Option A is correct.