An electric fan and a heater are marked as 10 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The power $P$ of an electrical appliance is related to its resistance $R$ and the operating voltage $V$ by the formula $P = \frac{V^2}{R}$.Since b

Vedclass · Accepted Answer

Less than that of the fan

Vedclass · Answer

(C) The power $P$ of an electrical appliance is related to its resistance $R$ and the operating voltage $V$ by the formula $P = \frac{V^2}{R}$.Since both appliances operate at the same voltage $V = 220 \ V$,the resistance is given by $R = \frac{V^2}{P}$.This implies that $R \propto \frac{1}{P}$,meaning resistance is inversely proportional to power.For the fan,$P_f = 100 \ W$,so $R_f = \frac{V^2}{100}$.For the heater,$P_h = 1000 \ W$,so $R_h = \frac{V^2}{1000}$.Comparing the two,since $1000 > 100$,it follows that $R_h Therefore,the resistance of the heater is less than that of the fan.

An electric fan and a heater are marked as $100 \ W, 220 \ V$ and $1000 \ W, 220 \ V$ respectively. The resistance of the heater is

Similar Questions

$n$ identical bulbs,each designed to draw a power $P$ from a certain voltage supply,are joined in series across that supply. The total power which they will draw is

$A$ constant voltage is applied between the two ends of a metallic wire. If both the length and the radius of the wire are doubled, the rate of heat developed in the wire

$A$ bulb has a specification of $1\, kW$ and $250\, V$. The resistance of the bulb is .................. $\Omega$.

The power dissipated by a uniform wire of resistance $100 \Omega$ when a potential difference of $120 \text{ V}$ is applied across its ends is (in $W$)

Heating coils rated at $220\, V$ are available with resistances of $55\,\Omega$,$110\,\Omega$,$220\,\Omega$,and $440\,\Omega$. Which resistance value will produce the maximum power?

Vedclass Test Series

Exam Paper Generator

Online Exam Module