Question

You are given 3 bulbs rated 180V - 40W, 240V - 60W, 200V - 100W. Which bulb has the highest resistance
2) 60W bulb
3) 100W
4) 1 & 2​

Answer 1

Given:-

• Three types of bulb is given .

1. $\tt Bulb\; A \:- 180V\: , \:40W.$
2. $\tt Bulb\; B -\: 240V\: , \;60W.$
3. $\tt Bulb \;C\: - 200V\; ,\: 100W .$

To Find:-

• $\tt Which\: bulb \:has\: a \:higher\: resistance.$

Formula Used:-

Power can be calculated using ,

$\large\purple{\underline{\boxed{\orange{\tt{\dag Power\:\:=\:\:\dfrac{Voltage^2}{Resistance}}}}}}$

Where ,

• V is voltage.
• R is resistance.

Calculation:-

$\underline{\tt{\pink{\hookrightarrow For\:Bulb\:A:-}}}$

• Voltage = 180V .
• Power = 40W .

So , on putting the values in the above formula stated we have :-

$\tt:\implies Power=\dfrac{V^2}{R}$

$\tt:\implies 40W =\dfrac{180V^2}{R}$

$\tt:\implies R =\dfrac{\cancel{180}\times\cancel{180}}{\cancel{40}}$

$\underline{\boxed{\red{\tt{\longmapsto R = 810\Omega}}}}$

_________________________________________

$\underline{\tt{\pink{\hookrightarrow For\:Bulb\:B:-}}}$

• Voltage = 240V .
• Power = 60W .

So , on putting the values in the above formula stated we have :-

$\tt:\implies Power=\dfrac{V^2}{R}$

$\tt:\implies 60W =\dfrac{240V^2}{R}$

$\tt:\implies R =\dfrac{\cancel{240}\times\cancel{240}}{\cancel{60}}$

$\underline{\boxed{\red{\tt{\longmapsto R = 960\Omega}}}}$

_________________________________________

$\underline{\tt{\pink{\hookrightarrow For\:Bulb\:C:-}}}$

• Voltage = 200V .
• Power = 100W .

So , on putting the values in the above formula stated we have :-

$\tt:\implies Power=\dfrac{V^2}{R}$

$\tt:\implies 100W =\dfrac{200V^2}{R}$

$\tt:\implies R =\dfrac{\cancel{200}\times\cancel{200}}{\cancel{100}}$

$\underline{\boxed{\red{\tt{\longmapsto R = 400\Omega}}}}$

$\tt{Hence\:here\:we\:can\:see\:that:}$

$\blue{\boxed{\boxed{\red{\bf{\dag R_B\:\:\:\:\:>R_A\:\:\:\:\:>R_C\:\:\:\:\:}}}}}$

$\underline\purple{\tt{\leadsto Hence\:60W\:bulb\:has\:largest\: resistance.}}$

Answer 2

Answer:

Option (2) 60 W bulb has the highest resistance  960Ω.

Explanation:

Let the bulbs be $B_1 ,B_2 and B_3$  and

Let the resistances be $R_1 ,R_2 and \ R_3$

Therefore given in the question that 3 bulbs rated  ,

Bulb one contain $B_1$ - 180V ,40W ,

Second bulb  contain $B_2$ - 240V ,60W and

Third bulb  contain $B_3$ - 200V , 100W

As we know that , Power = $\frac{voltage ^2}{Resistance }$

Therefore , For bulb $B_1$ ,

Given , voltage = 180V and Power = 40 W .

⇒$Power = \frac{V^2}{R}$

⇒40 = $\frac{180 .180 }{R}$ ⇒ $R_1$ = $\frac{32400}{40 }$ = 810 Ω

Similarly , for  the second bulb $B_2$ ;

Given , Voltage = 240V and power = 60 W .

⇒60 = $\frac{240 .240}{R}$ ⇒$R_2$ = $\frac{240.240}{60}$ = $\frac{57600}{60}$ = 960 Ω

For the third bulb $B_3$ ;

Given voltage = 200V and power = 100W .

⇒Power  = $\frac{V^2}{R}$

⇒100 = $\frac{200 .200}{R}$  ⇒$R_3$ = $\frac{200 . 200}{100}$ = 400Ω

Therefore , here we can see that , second bulb has the highest resistance 960Ω .

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