Question

A current of 1 A flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, What change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place ? Give reason. Draw Circuit diagram

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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the Concept of Parallel and Series combination of the resistances has been used. We see we are given two cases. In first case, we need to find the resistance of the bulb. This we can do by using Ohm's Law we can find that. Then for second case, we need to find the change in potential difference across different parts.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{R_{eq_{(s)}}\;=\;\bf{R_{1}\;+\;R_{2}\;+.....\;+\;R_{n}}}}}$

$\\\;\boxed{\sf{\pink{\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{R_{1}}\;+\;\dfrac{1}{R_{2}}\;+.....\;+\;\dfrac{1}{R_{n}}}}}}$

$\\\;\boxed{\sf{\pink{R_{eq}\;=\;\bf{\dfrac{V}{I}}}}}$

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★ Solution :-

~ Case I (first attachment) ::

Given,

» Current through the circuit = I = 1 A

» Resistance of the conductor = R₂ = 5 Ω

» Resistance of Bulb = R₁

» Potential difference of the circuit = V = 10 Volts

Then total resistance of the circuit :-

$\\\;\sf{:\rightarrow\;\;R_{eq_{(s)}}\;=\;\bf{R_{1}\;+\;R_{2}\;+.....\;+\;R_{n}}}$

This shows the net equivalent resistance of a Series Combination. Then,

$\\\;\bf{:\rightarrow\;\;R_{eq_{(s)}}\;=\;\bf{\green{(R_{1}\;+\;5)\;\;\Omega}}}$

This is the effective resistance of the circuit ↑ .

Then using Ohm's Law, we get,

$\\\;\sf{:\rightarrow\;\;R_{eq}\;=\;\bf{\dfrac{V}{I}}}$

By applying values, we get,

$\\\;\sf{:\Longrightarrow\;\;(R_{1}\;+\;5)\;=\;\bf{\dfrac{10}{1}}}$

$\\\;\sf{:\Longrightarrow\;\;R_{1}\;+\;5\;=\;\bf{10}}$

$\\\;\sf{:\Longrightarrow\;\;R_{1}\;\;=\;\bf{10\;-\;5}}$

$\\\;\bf{:\Longrightarrow\;\;R_{1}\;\;=\;\bf{\blue{5\;\;\Omega}}}$

✒ Resistance of Bulb = 5Ω

$\\\;\underline{\boxed{\tt{Resistance\;\;of\;\;Bulb\;=\;\bf{\purple{5\;\;\Omega}}}}}$

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~ Case II (second attachment) ::

Given,

» 10 Ω resistance is connected in parallel with series combination.

» R₃ = 10 Ω

Then, this total effective resistance will be given as,

$\\\;\sf{:\rightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{R_{1}}\;+\;\dfrac{1}{R_{2}}\;+.....\;+\;\dfrac{1}{R_{n}}}}$

Since, R₁ and R₂ are in series combination, then

$\\\;\sf{:\rightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{R_{1}\;+\;R_{2}}\;+\;\dfrac{1}{R_{3}}}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{5\;+\;5}\;+\;\dfrac{1}{10}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{10}\;+\;\dfrac{1}{10}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{2}{10}}}$

$\\\;\sf{:\Longrightarrow\;\;\dfrac{1}{R_{eq_{(p)}}}\;=\;\bf{\dfrac{1}{5}}}$

Now taking the reciprocal, we get

$\\\;\bf{:\Longrightarrow\;\;R_{eq_{(p)}}\;=\;\bf{\red{5\;\;\Omega}}}$

Hence, now the total resistance of the circuit is 5 Ω .

• Rule 1 (for current) : In parallel combination the current gets divided between the components while in series it remains same throughout each component.

• Rule 1 (for potential difference) : In parallel combination the potential difference remains same across each component while in series combination the potential difference gets divided between the components.

Now we know that Potential Difference remains same in each component. So,

• Potential Difference = V = 10 Volts

• Let the current through the circuit be I

Then using Ohm's law,

$\\\;\sf{:\rightarrow\;\;R_{eq}\;=\;\bf{\dfrac{V}{I}}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;5\;=\;\bf{\dfrac{10}{I}}}$

$\\\;\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{10}{5}}}$

$\\\;\sf{:\Longrightarrow\;\;I\;=\;\bf{\orange{2\;\;A}}}$

Hence, now current through circuit will be 2A .

• This means 1 A of current flows through the 10Ω resistor and 1 A current flows through series combination of R₁ and R₂. Since in series combination current remains same at each and every component. So current through the conductor will be 1 A .

Hence, there is no change in current through the conductor.

• Then we see that (bulb and conductor) are connected in parallel with the resistor of 10Ω . This means 10 V potential difference remains across 10Ω resistor and across series combination. And we see that both component that is bulb and conductor are of same resistance. Then, according to Ohm's law, Potential difference at bulb will be (5 × 1) that is 5 V. This is also same as first case because there also bulb and conductor were in series only.

Hence there is no change in Potential Difference across the Bulb.

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★ More to know :-

$\\\;\sf{\gray{\leadsto\;\;Power\;=\;V\;\times\;I}}$

$\\\;\sf{\gray{\leadsto\;\;Heat\;=\;V\;\times\;I\;\times\;T}}$

$\\\;\sf{\gray{\leadsto\;\;Heat\;=\;P\;\times\;T}}$

*Note :: Here Ømega shows Ω (for app users).