Question

An electric bulb of resistance 500 2 draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.​

Answer 1

Given:

• Resistance, R = 500 Ω
• Current, I = 0.4 A

To find:

• Power, P
• Potential Difference, V

Solution:

(b) Firstly, calculating the potential difference using ohm's law,

• V = IR

➞ V = 0.4 × 500

➞ V = 200V

Hence, potential difference between at its end is 200 volts.

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(a) Now, calculating the power of the bulb,

As we know that,

• P = VI

➞ P = 200 × 0.4

➞ P = 20 × 4

➞ P = 80 W

Hence, the power of the bulb is 80 watt.

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Answer 2

$\frak{Given}\begin{cases}\sf{\;Resistance\;of\;electric\;bulb={\bf{500\;Ω}}}\\\\\sf{\;Current\;drawn\;from\;the\;source={\bf{0.4\;A}}}\end{cases}$

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Need to find: The power of the bulb & the potential difference at its end.

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★ Calculate the potential difference at its end, using ohm's law,

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$\star\;\boxed{\pink{\sf{V=IR}}}$

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where,

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• V is the potential difference, I is the current & R is the resistance of the electric bulb, therefore,

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$:\implies\sf{V=0.4\times 500}\\\\\\:\implies{\underline{\boxed{\purple{\frak{V=200\;V}}}}}\;\bigstar$

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$\therefore\;{\underline{\sf{Hence,\;the\;</p><p>potential\;is\;{\textsf{\textbf{200\;volts}}}}.}}$

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★ Calculating the power of the bulb,

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$\underline{\bf{\dag}\frak{\;As\;we\;know\;that\;:}}$

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$\star\;\boxed{\pink{\sf{P=VI}}}$

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where,

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• P is the power of the bulb, V is the potential difference & I is the current, therefore,

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$:\implies\sf{P=200\times 0.4}\\\\\\:\implies\sf{P=20\times 4}\\\\\\:\implies{\underline{\boxed{\purple{\frak{P=80\;W}}}}}\;\bigstar$

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$\therefore\;{\underline{\sf{Hence,\;the\;</p><p>power\;of\;the\;bulb\;is\;{\textsf{\textbf{80\;Watt}}}}.}}$

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