Question

An electric bulb consumes 6000 J of energy in one minute. If the potential difference of mains is 200 volts, find the resistance of its filament​

Answer 1

Answer:

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The resistance of its filament is 400 Ω.

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Explanation:

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Given:

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• Electric work (W) = 6000 J
• Potential difference (V) = 200 V
• Time (t) = 1 min = 60 s.

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To find:

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Resistance (R) of its filament.

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Solution:

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We know that,

W = V$^{2}$ /R × t

Now substitute the known values

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⇒ $6000 = \frac{200 \times 200 \times 60}{6000}$

⇒ $R = \frac{200 \times 200 \times 60}{6000}$

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Therefore, R = 400 Ω.

Answer 2

Explanation:

200×200×600÷600

=400