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

Explanation:

Energy (power×time) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.

Energy dissipated = Pt or VI×tasP=VI.

In the question, it is given that the current I is 0.5 A and voltage V is 4 V. Thus the power dissipated is calculated as P=VI=4V×0.5A=2W.

Substituting V=IR (Ohm's law) in the above formula, we get, P=I

2

×R.

Therefore, the total resistance in the circuit is calculated using the relation R=

I

2

P

=

0.5

2

2

=8ohms.

The internal resistance of the cell is 2.5 ohms, hence, the resistance of the bulb is 8−2.5=5.5 ohms.

The power dissipated by the bulb in 10 minutes is calculated from the formula P=I

2

×R×t, where I=0.5A,R=5.5 ohms and t=10 minutes, that is 600 seconds.

Hence, P=0.5

2

×5.5×600=825J.

Answer 2

Answer:

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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

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Solution:

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We know that,

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W = V$^{2}$ / R × t

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

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⟹ R = (200*200*60) ÷ 6000

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= 400 Ω

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Thus, resistance of its filament is 400 Ω.