Question

Two bulbs R and S are marked 60W, 220V and 60W, 110V respectively. Find out the
ratio of their resistances

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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Power and Resistance relationship has been used. We see that we are given the Power and Potential Difference of two bulbs. The question is that that what will be the ratio of their resistances. We can calculate it by calculating the resistances of each bulb and then taking their ratio.

Let's do it !!

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★ Formula Used

$\\\;\boxed{\sf{\pink{Power\;=\;\bf{\dfrac{V^{2}}{R}}}}}$

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★ Solution :-

Given,

» Power of bulb R = P₁ = 60 W

» Potential Difference of bulb R = V₁ = 220 V

» Power of bulb S = P₂ = 60 W

» Potential Difference of bulb S = V₂ = 110 V

• Let the Resistance of bulb R be R₁

• Let the Resistance of bulb S be R₂

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~ For the Resistance of bulb R ::

By Power - Resistance relationship, we know that

$\\\;\sf{\rightarrow\;\;Power\;=\;\bf{\dfrac{V^{2}}{R}}}$

• Here V is the Potential Difference and R is the resistance.

By applying values we get,

$\\\;\sf{\rightarrow\;\;P_{1}\;=\;\bf{\dfrac{V_{1} ^{2}}{R_{1}}}}$

$\\\;\sf{\Longrightarrow\;\;R_{1}\;=\;\bf{\dfrac{V_{1} ^{2}}{P_{1}}}}$

$\\\;\sf{\Longrightarrow\;\;R_{1}\;=\;\bf{\dfrac{(220)^{2}}{60}}}$

$\\\;\bf{\Longrightarrow\;\;R_{1}\;=\;\bf{\green{\dfrac{220\:\times\:220}{60}\;\Omega}}}$

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~ For the Resistance of bulb S ::

By Power - Resistance relationship, we know that

$\\\;\sf{\rightarrow\;\;Power\;=\;\bf{\dfrac{V^{2}}{R}}}$

• Here V is the Potential Difference and R is the resistance.

By applying values, we get

$\\\;\sf{\rightarrow\;\;P_{2}\;=\;\bf{\dfrac{V_{2} ^{2}}{R_{2}}}}$

$\\\;\sf{\Longrightarrow\;\;R_{2}\;=\;\bf{\dfrac{V_{2} ^{2}}{P_{1}}}}$

$\\\;\sf{\Longrightarrow\;\;R_{2}\;=\;\bf{\dfrac{(110)^{2}}{60}}}$

$\\\;\bf{\Longrightarrow\;\;R_{2}\;=\;\bf{\orange{\dfrac{110\:\times\:110}{60}\;\Omega}}}$

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~ For the Ratio of Resistances ::

This is given as,

$\\\;\bf{\mapsto\;\;\red{Ratio\;of\;Resistances}\;=\;\bf{\dfrac{\green{R_{1}}}{\orange{R_{2}}}}}$

By applying values, we get

$\\\;\sf{\mapsto\;\;\dfrac{R_{1}}{R_{2}}\;=\;\bf{\dfrac{\bigg(\dfrac{220\:\times\:220}{60}\bigg)}{\bigg(\dfrac{110\:\times\:110}{60}\bigg)}}}$

Cancelling the denominators, we get

$\\\;\sf{\mapsto\;\;\dfrac{R_{1}}{R_{2}}\;=\;\bf{\dfrac{220\:\times\:220}{110\:\times\:110}}}$

$\\\;\sf{\mapsto\;\;\dfrac{R_{1}}{R_{2}}\;=\;\bf{\dfrac{2\:\times\:2}{1\:\times\:1}}}$

$\\\;\bf{\mapsto\;\;\dfrac{R_{1}}{R_{2}}\;=\;\bf{\blue{\dfrac{4}{1}}}}$

In the ratio form, this can be written as

$\\\;\bf{\blue{\mapsto\;\;R_{1}\;:\;R_{2}\;=\;\bf{4\;:\;1}}}$

$\\\;\underline{\boxed{\tt{Ratio\;\:of\;\:Resistances\;=\;\bf{\purple{4\;:\;1}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;I\;=\;\dfrac{Q}{T}}$

$\\\;\sf{\leadsto\;\;V\;=\;\dfrac{W}{Q}}$

$\\\;\sf{\leadsto\;\;P\;=\;V\;\times\;I}$

$\\\;\sf{\leadsto\;\;H\;=\;P\:\times\:T}$

$\\\;\sf{\leadsto\;\;H\;=\;I^{2}\:\times\:R\:\times\:T}$

Answer 2

Answer:

Refer to the attachment mate..

Hope it would help you..