Question

Solve this don't solve if u don't know ​

Answer 1

★ Concept :-

Here the concept of Ohm's Law has been used. Firstly we will calculate the equivalent resistance of different parts and then combine them to get total resistance.

★ Formula Used :-

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

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

$\;\boxed{\sf{\pink{R\;=\;\bf{\dfrac{Votage}{I}}}}}$

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★ Solution :-

Given,

» Potential Difference of circuit = Voltage = 28 V

From attached image we see, that we have named different resistances. Let's look at it.

» Resistance a = 5 Ω

» Resistance b = 3 Ω

» Resistance c = 3 Ω

» Resistance d = 4 Ω

» Resistance e = 3 Ω

» Resistance f = 2 Ω

» Resistance g = 4 Ω

» Resistance h = 10 Ω

» Resistance i = 10 Ω

» First area of circuit = ADEF

» Second area of circuit = ABCD

» Third area of circuit = CBGH

• Let the current in the circuit be 'I'

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~ For the Equivalent Resistance of Part ADFE ::

Here we see that equivalent resistances we have to calculate the equivalent resistance of c, d and e in series which is in parallel with resistance h.

Then,

• For series we know that,

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

Now applying values, we get

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{c\:+\:d\:+\:e}}$

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{3\:+\:4\:+\:3}}$

$\;\bf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{\green{10\;\;\Omega}}}$

• Now for parallel combination,

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

$\;\sf{\rightarrow\;\;\dfrac{1}{R_{eq_{p}}}\;=\;\bf{\dfrac{1}{h}\;+\;\dfrac{1}{10}}}$

$\;\sf{\rightarrow\;\;\dfrac{1}{R_{eq_{p}}}\;=\;\bf{\dfrac{1}{10}\;+\;\dfrac{1}{10}}}$

$\;\sf{\rightarrow\;\;\dfrac{1}{R_{eq_{p}}}\;=\;\bf{\dfrac{2}{10}}}$

$\;\sf{\rightarrow\;\;\dfrac{1}{R_{eq_{p}}}\;=\;\bf{\dfrac{1}{5}}}$

$\;\bf{\rightarrow\;\;R_{eq_{p}}\;=\;\bf{\orange{5}}}$

This is the equivalent resistance of part ADF that is 5 Ω

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~ For the Equivalent Resistance of Part ABCD ::

For this firstly we will calculate the resistance in series combination of resistances b, f and equivalent resistance of AD. Now this series combination will be in parallel with resistance i.

• For series, we know that

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

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{b\;+\;f\;+\;ADFE_{(eq)}}}$

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{3\:+\:2\:+\:5}}$

$\;\bf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{\blue{10\;\;\Omega}}}$

• For parallel combination,

$\;\sf{\Longrightarrow\;\;\dfrac{1}{R_{eq_{p}}}\;=\;\bf{\dfrac{1}{R_{1}}\;+\;\dfrac{1}{R_{2}}\;+\;...\;+\;\dfrac{1}{R_{n}}}}$

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

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

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

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

$\;\bf{\Longrightarrow\;\;R_{eq_{p}}\;=\;\bf{\red{5}}}$

This is the equivalent resistance of part ABCD that is 5 Ω

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~ For the Equivalent Resistance of Part CBGH ::

For this we will calculate the equivalent resistance in series combination of resistance of resistance a, g and equivalent resistance of part ABCD.

Then,

• For series combination,

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

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{a\:+\:g\:+\:ABCD_{(eq)}}}$

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{5\:+\:4\:+\:5}}$

$\;\sf{\rightarrow\;\;R_{eq_{s}}\;=\;\bf{\pink{14\;\;\Omega}}}$

Since, we calculated this resistance by adding all other equivalent resistance of different parts of circuit, so this is the equivalent resistance of the circuit.

$\;\bf{\leadsto\;\;Equivalent\;Resistance\;of\;Circuit\;=\;R\;=\;\pink{14\;\;\Omega}}$

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~ For Total Current in the Circuit ::

By Ohm's Law, we know that

$\;\sf{\mapsto\;\;R\;=\;\bf{\dfrac{Votage}{I}}}$

By applying values, we get

$\;\sf{\mapsto\;\;14\;=\;\bf{\dfrac{28}{I}}}$

$\;\sf{\mapsto\;\;I\;=\;\bf{\dfrac{28}{14}}}$

$\;\sf{\mapsto\;\;I\;=\;\bf{2\;\;Ampere}}$

We know that, the 5 Ω resistance is connected in first position with the battery. This means the current flowing through the 5 Ω resistance will be equal to the total current of circuit.

$\;\underline{\boxed{\tt{Current\;\:through\;\:5\:\Omega\;\:resistance\;=\;\bf{\purple{2\;\;Ampere}}}}}$

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Answer 2

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