What is the resistance of a 40 W lamp which is lighted at full brilliance by a current of ( text1/ text3) textA ?

Question

What is the resistance of a 40 W lamp which is lighted at full brilliance by a current of ( text1/ text3) textA ? (A)  450 Ω  (B)  360 Ω  (C)  120 Ω  (D)  13.33 Ω

Tardigrade Admin · Accepted Answer

Power of electric bulb is given by P=(V2/R) But from Ohm's law, V = i R So, P=((i R)2/R)=i2 R ∴ R=(p/i2) Here, P=40 W , i=(1/3) A Hence, R=(40/(1 / 3)2)=40 × 9=360 Ω

Q. What is the resistance of a $40 W$ lamp which is lighted at full brilliance by a current of $\frac{1}{3} A$ ?

$450 Ω$

$360 Ω$

$120 Ω$

$13.33 Ω$

Power of electric bulb is given by
$P = \frac{V ^{2}}{R}$
But from Ohm's law,
$V = i R$
So, $P = \frac{( i R ) ^{2}}{R} = i^{2} R$
$∴ R = \frac{p}{i ^{2}}$
Here, $P = 40 W, i = \frac{1}{3} A$
Hence, $R = \frac{40}{( 1/3 ) ^{2}} = 40 \times 9 = 360 Ω$

Q. What is the resistance of a 40W lamp which is lighted at full brilliance by a current of 31​A ?

450Ω

360Ω

120Ω

13.33Ω