Question

If a current of 0.5 A flows in a 75 W light bulb when the voltage difference between the ends of the filament is 150 V, what is the resistance of the filament?

300 Ω
75 Ω
2 Ω

Answer 1

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Given,

voltage v = 150 V

current I = 0.5 A

power P = 75 W

resistance R = to find

Method 1:

we know that,

V = IR

$\implies \: R = \frac{V }{I} \\ = \frac{150}{0.5} \\ = 300 \: \: ohm$

Method 2:

$P = \frac{V^2}{R} \\ \implies \:R</p><p> = \frac{V^2}{P} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ ({150)}^{2} }{75} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 300 \: \: ohm</p><p>$

Answer 2

Answer:

Given :

➡ Voltage (V) = 150 V

➡ Current (I) = 0.5 A

➡ Power (P) = 75 W

➡ Resistance (R) = to find

Method 1 :-

we know that,

V = IR

➡ 150 = (0.5) x R

➡ R = 150/0.5

➡ R = 1500/05

➡ R = 300 ohms

Method 2 :-

We know that :

P = V²/R

➡ 75 = (150)²/R

➡ 75 = 22500/R

➡ 75/22500 = 1/R

➡ 1/300 = 1/R

➡ R = 300 ohms

Hope it helps you !!