Question

what will be the resistance of 240V-60W lamp:-

a)480 ohm b) 960 ohm

c)240 ohm d)720 ohm​

Answer 1

Answer:

Given

the lamp is rated 240V-60W

According to joule's law of heating

We know

$power = \frac{(potential \: difference) {}^{2} }{resistance} \\ p = \frac{v {}^{2} }{r}$

now here

p= 60W, and v = 240V

So

$p = \frac{v {}^{2} }{r} \\ 60 = \frac{(240) {}^{2} }{r} \\ r = \frac{(240) {}^{2} }{60} \\ r = \frac{240 \times 240}{60} \\ r = 4 \times 240 \\ r = 960 \: ohm$

So option b) 960 ohm is correct answer.

I hope it will help.

Answer 2

Answer:

The lamp has a working resistance of 960 ohms.

Explanation:

Given

the lamp is rated 240V-60W

According to joule's law of heating

We know

\begin{gathered}power = \frac{(potential \: difference) {}^{2} }{resistance} \\ p = \frac{v {}^{2} }{r} \end{gathered}

power=

resistance

(potentialdifference)

2

p=

r

v

2

now here

p= 60W, and v = 240V

So

\begin{gathered}p = \frac{v {}^{2} }{r} \\ 60 = \frac{(240) {}^{2} }{r} \\ r = \frac{(240) {}^{2} }{60} \\ r = \frac{240 \times 240}{60} \\ r = 4 \times 240 \\ r = 960 \: ohm\end{gathered}

p=

r

v

2

60=

r

(240)

2

r=

60

(240)

2

r=

60

240×240

r=4×240

r=960ohm

So option b) 960 ohm is correct answer.

I hope it will help.

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