Question

A 50 W television and 3 kW oven are connected to a 300 V supply. The resistance of each appliance is respectively

Answer 1

Given ,

Electric power of television = 50 w

Electric power of oven = 3 kw or 3000 w

Potential difference = 300 v

As we know that , the electric power is given by

$\star \: \: \large{ \fbox{\sf P = \frac{ {(v)}^{2} }{r} }}</p><p>$

Thus ,

The resistance of television is

$\Rightarrow \sf R = \frac{ {(300)}^{2} }{50} \\ \\ </p><p>\Rightarrow \sf R = \frac{90000}{50} \\ \\ \Rightarrow \sf </p><p>R = 1800 \: Ω$

The resistance of oven is

$\Rightarrow \sf R = \frac{ {(300)}^{2} }{3000} \\ \\ \Rightarrow \sf</p><p>R = \frac{90000}{3000} \\ \\ \Rightarrow \sf</p><p>R = 30 \: Ω$

$\therefore \bold{\sf \underline{The \: resistance \: of \: television \: and \: oven \: are \: 1800 \: Ω \: and \: 30 \: Ω</p><p>}}$

Answer 2

$\rule{200}2$

$\huge\bold{\mathtt{QUESTION}}$

➡ A 50 W television and 3 kW oven are connected to a 300 V supply. The resistance of each appliance respectively is ?

$\rule{200}2$

$\huge\bold{\mathtt{SOLUTION}}$

⚡ Given -

↪ Electric power of television = 50 watt

↪ Electric power of oven = 3 Kw = 3000 watt

↪ Potential Difference = 300 volt

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⚡ To Find -

↪ Resistance of each appliance

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☯ Formula to be used -

$\sf \:\:\:\:\:\:\:\:\:\:\:Power\:=\: \dfrac{{(Potential \:Difference)}^{2}}{Resistance}$

▶So , $\sf \:\:\:\:\:\:\:\:\:\:\:Resistance\:=\: \dfrac{{(Potential \:Difference)}^{2}}{Power}$

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✳ Resistance of television -

$\sf \:\:\:\:\:\:\:\:\:\:\:Resistance\:=\: \dfrac{{(200)}^{2}}{50}$

$\sf \:\:\:\:\:\:\:\:\:\:\:{\underline{\red{Resistance\:=\: 1800 \:\Omega}}}$

✳ Resistance of oven -

$\sf \:\:\:\:\:\:\:\:\:\:\:Resistance\:=\: \dfrac{{(300)}^{2}}{3000}$

$\sf \:\:\:\:\:\:\:\:\:\:\:{\underline{\red{Resistance\:=\: 30 \:\Omega}}}$

$\rule{200}2$