Question

7. The resistance of hot filament of the bulb is about 10 times the cold resistance.

What will be the resistance of 100W – 220V lamp, when not in use?

(a) 48Ω (b) 400Ω (c) 484Ω (d) 48.4Ω​

Answer 1

Answer: 48.4 Ω

Explanation:

P = V²/R

=>R = V²/P

=>R = (220)² / 100 = 484Ω

This is the resistance of hot filament ,  

we are asked about the resistance when lamp is not in use , i.e , Cold resistance.

therefore ,

R(cold) x 10 = R (hot)

=> R(cold) = 484/10 Ω  = 48.4 Ω

Hope it helps

Please mark me brainliest

Answer 2

$\bigstar$ Given

The resistance of hot filament of the bulb is about 10 times the cold resistance

Power = 100 W

Voltage = 220 V

$\bigstar$ To Find

What will be the resistance of lamp when not in use

$\bigstar$ Solution

We know that

$\red{\rm \longrightarrow R=\dfrac{V^{2}}{P}}$

Here

• R = Resistance
• V = Voltage
• P = Power

Units

• R = Ohm (Ω)
• V = Volt (V)
• P = Wattage (Watt)

Note:

1. When in use = hot
2. When not in use = cold

According to the question

We are asked to find resistance of lamp when not in use (cold)

Therefore

We must find $\rm " \ R_{cold} \ "$

Given that

Power = 100 W

Voltage = 220 V

Hence

• P = 100 W
• V = 220 V

Substituting the values

Calculating $\rm " \ R_{hot} \ "$

We get

$\rm \longrightarrow R_{hot}=\dfrac{220^{2}}{100} \ \Omega$

$\rm \longrightarrow R_{hot}=\dfrac{220 \times 220}{100} \ \Omega$

$\rm \longrightarrow R_{hot}=\dfrac{48400}{100} \ \Omega$

Therefore

$\orange{\rm \longrightarrow R_{hot}=484 \ \Omega}$

Given that

The resistance of hot filament of the bulb is about 10 times the cold resistance

Hence

$\pink{\rm R_{hot}=10 \times R_{cold}}$

Therefore

$\rm \implies R_{cold}=\dfrac{R_{hot}}{10}$

So

$\rm \implies R_{cold}=\dfrac{484}{10} \ \Omega$

Therefore

$\purple{\rm \implies R_{cold}=48.4 \ \Omega}$

Hence

$\checkmark$ Option (d) is correct