Question

A bulb has voltage rating of 220V and power rating of 40W

(a) Calculate the resistance of the bulb.

(b) What will happen if a voltage source of 200V is connected across the bulb?

(c) How can the bulb be made to glow with normal brightness if a voltage of emf 300V is available?​

Answer 1

Explanation:

220 V / 40 W rating mean that the bulb will dissipate power equal to 40 W when the voltage of 220 V is applied across it . The brightness of a bulb is proportional to the power dissipated across it . A power dissipation of 40 W for this bulb will make it glow with normal brightness.

$(a) \: Resistance \: of \: bulb \: </p><p>, R = \frac{ {voltage \: rating}^{2} }{power \: rating} \\ \\ ==> R = \frac{220 \times 220}{40} = 1210 \: ohm$

(b) If the voltage across bulb is 200 V

$Power \: dissipated \: = \frac{221 \times 220}{1210} = 33.05$

As the power dissipated is less than 40 W , the bulb will glow less than the normal brightness .

(c) If a voltage greater than the voltage rating ( = 220 V ) is applied , the bulb will glow more brightly , but it can be damaged . Hence for a voltage supply of 330 V , we will have to divert the voltage across resistance S connected in series with the bulb .

S should be selected so that potential difference across the bulb is 220 V .

$==> 220 = 330( \frac{R}{R + S}) \\ \\ ==> \: 220 = 330( \frac{1210}{1210 +S} ) \\ \\ ==> \frac{2}{3} = \frac{1210}{1210 +S} \\ \\==> 1210 +S = \frac{1210 \times 3}{2} \\ \\ ==>1210 +S = 1815 \\ \\==>S = 1815 - 1210 = 605 \: ohm$

==> S = 605 ohm

Answer 2

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$\huge {\purple {\fbox {\bigstar {\mathbf {\red {ur\: question}}}}}}$

$\huge {\mathbf {\purple {Q.}}}}$A bulb has voltage rating of 220V and power rating of 40W

(a) Calculate the resistance of the bulb.

(b) What will happen if a voltage source of 200V is connected across the bulb?

(c) How can the bulb be made to glow with normal brightness if a voltage of emf 300V is available?

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$\huge {\purple {\fbox {\bigstar {\mathbf {\red {ur\: answer✓}}}}}}$

$\frac{220v}{40w}$ʀᴀᴛɪɴɢ ᴍᴇᴀɴ ᴛʜᴀᴛ ᴛʜᴇ ʙᴜʟʙ ᴡɪʟʟ ᴅɪssɪᴘᴀᴛᴇ ᴘᴏᴡᴇʀ ᴇϙᴜᴀʟ ᴛᴏ 40 ᴡ ᴡʜᴇɴ ᴛʜᴇ ᴠᴏʟᴛᴀɢᴇ ᴏғ 220 ᴠ ɪs ᴀᴘᴘʟɪᴇᴅ ᴀᴄʀᴏss ɪᴛ . ᴛʜᴇ ʙʀɪɢʜᴛɴᴇss ᴏғ ᴀ ʙᴜʟʙ ɪs ᴘʀᴏᴘᴏʀᴛɪᴏɴᴀʟ ᴛᴏ ᴛʜᴇ ᴘᴏᴡᴇʀ ᴅɪssɪᴘᴀᴛᴇᴅ ᴀᴄʀᴏss ɪᴛ . ᴀ ᴘᴏᴡᴇʀ ᴅɪssɪᴘᴀᴛɪᴏɴ ᴏғ 40 ᴡ ғᴏʀ ᴛʜɪs ʙᴜʟʙ ᴡɪʟʟ ᴍᴀᴋᴇ ɪᴛ ɢʟᴏᴡ ᴡɪᴛʜ ɴᴏʀᴍᴀʟ ʙʀɪɢʜᴛɴᴇss.

$\begin{lgathered}(a) \: Resistance \: of \: bulb \: , R = \frac{ {voltage \: rating}^{2} }{power \: rating} \\ \\ \implies R = \frac{220 \times 220}{40} = 1210 \: ohm \\ \\\end{lgathered}$

$\sf (b) If \:the \:voltage\: across\: bulb\: is \:200 V$

$Power \: dissipated \: = \frac{221 \times 220}{1210}$

$\sf As\:the\: power\: dissipated \:is \:less\: than\: 40 W ,$

$\sf the\: bulb\: will\: glow\: less\: than\: the \:normal\: brightness$

(ᴄ) ɪғ ᴀ ᴠᴏʟᴛᴀɢᴇ ɢʀᴇᴀᴛᴇʀ ᴛʜᴀɴ ᴛʜᴇ ᴠᴏʟᴛᴀɢᴇ ʀᴀᴛɪɴɢ ( = 220 ᴠ ) ɪs ᴀᴘᴘʟɪᴇᴅ , ᴛʜᴇ ʙᴜʟʙ ᴡɪʟʟ ɢʟᴏᴡ ᴍᴏʀᴇ ʙʀɪɢʜᴛʟʏ , ʙᴜᴛ ɪᴛ ᴄᴀɴ ʙᴇ ᴅᴀᴍᴀɢᴇᴅ . ʜᴇɴᴄᴇ ғᴏʀ ᴀ ᴠᴏʟᴛᴀɢᴇ sᴜᴘᴘʟʏ ᴏғ 330 ᴠ , ᴡᴇ ᴡɪʟʟ ʜᴀᴠᴇ ᴛᴏ ᴅɪᴠᴇʀᴛ ᴛʜᴇ ᴠᴏʟᴛᴀɢᴇ ᴀᴄʀᴏss ʀᴇsɪsᴛᴀɴᴄᴇ s ᴄᴏɴɴᴇᴄᴛᴇᴅ ɪɴ sᴇʀɪᴇs ᴡɪᴛʜ ᴛʜᴇ ʙᴜʟʙ .

$\sf S\: should \:be\: selected\: so \:that\: potential\: difference \:across\: the\: bulb \:is \:220 V$

$\begin{lgathered}==> 220 = 330( \frac{R}{R + S}) \\ \\ \implies \: 220 = 330( \frac{1210}{1210 +S} ) \\ \\ \implies \frac{2}{3} = \frac{1210}{1210 +S} \\ \\ \implies 1210 +S = \frac{1210 \times 3}{2} \\ \\ \implies 1210 +S = 1815 \\ \\ \implies S = 1815 - 1210 = 605 \: ohm\end{lgathered}$

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