Question

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Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

★ Concept :-

Here the concept of Resistance has been used. We see that we are given the diagram of a circuit part according to the qúestion. So firstly we can make the real circuit part from the given and them we can make simplification of that circuit. Finally then we can derive certain values by using Trignometry and Geometry. And then we can apply it in the formula and find the total resistance.

Let's do it !!

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★ Solution :-

Given,

• » Letter constructed = A
• » Resistance per cm = 1 Ω per cm
• » Type of wire = Uniform Wire
• » Length of each side of the Letter = 20 cm
• » Length of cross piece in middle of letter = 10 cm

From Figure :: (Attachment I)

>> RP = RS = 20 cm

(since it's given that length of each side of letter = 20 cm)

>> QT = 10 cm

(since it's given that length of middle piece = 10 cm)

Let RQ be 'y' cm

Since, RP = RS

So,

>> RQ = RT = y cm

(since they are part of sides with equal length)

Since total RP = 20 cm and RQ = y cm . So,

>> PQ = ST = (20 - y) cm

(since RP = RS and RQ = RT)

• For the Apex Angle ::

Here the Apex Angle is ∠QRT .

We see that RQ = RT. And since it's letter A, thus ∆QRT will be an equilateral triangle. And all the angles of equilateral triangle are equal to 60° .

So,

• • ∠QRT = 60 °

• For the value of y ::

Let's use construction in order to find y. Draw a perpendicular bisector from R to QT which meéts QT at A in ∆QRT.

This will give us two triangles where,

→ ∠QRA = ∠TRA = 30°

→ QA = AT = 5 cm

→ RQ = RT = y cm

→ ∠RAQ = ∠RAT = 90°

So ∆RAQ and ∆RAT are right angled triangles.

Now let's consider ∆QRA .

Here,

$\;\tt{\leadsto\;\;\sin 30^{\circ}\;=\;\dfrac{Perpendicular}{Hypotenuse}\;=\;\dfrac{QA}{RQ}}$

sin 30° = ½

By applying values, we get

$\;\tt{\leadsto\;\;\dfrac{1}{2}\;=\;\dfrac{5}{y}}$

$\;\bf{\leadsto\;\;\orange{y\;=\;10\:\;cm}}$

• • For the value of resistance of different parts ::

Firstly we shall find the length of each part.

>> RQ = RT = y = 10 cm

>> PQ = ST = (20 - y) cm = (20 - 10) cm = 10 cm

>> QT = 10 cm

We are given that resistance per cm is 1 Ω .

::: Resistance of RQ = Resistance of RT = 10 × 1 = 10 Ω

::: Resistance of QT = 10 × 1 = 10 Ω

:: Resistance of PQ = Resistance of ST = 10 × 1 = 10 Ω

• Let the resistance of PQ be R₁
• Let the resistance of RQ be R₂
• Let the resistance of RT be R₃
• Let the resistance of QT be R₄
• Let the resistance of ST be R₅

• For the equivalent Resistance ::

(Refer the second attachment which is simplification of the circuit for convenience)

• We see that, R₁ and R₂ are in series with each other. So, for series combination equivalent resistance is given as,

$\;\sf{\rightarrow\;\;R_{eq}\;=\;R_{1}\;+\;R_{2}}$

By applying values, we get

$\;\tt{\rightarrow\;\;R_{eq}\;=\;10\;+\;10}$

$\;\bf{\rightarrow\;\;\green{R_{eq}\;=\;20\;\text{\O}mega}}$

• • We see that R₃ is in parallel combination with the combination of R₁ and R₂ . So, for parallel combination equivalent resistance is given as,

$\;\sf{\rightarrow\;\;\dfrac{1}{R_{eq}}\;=\;\dfrac{1}{R_{1\:and\:2}}\;+\;\dfrac{1}{R_{3}}}$

By applying values, we get

$\;\tt{\rightarrow\;\;\dfrac{1}{R_{eq}}\;=\;\dfrac{1}{20}\;+\;\dfrac{1}{10}}$

$\;\tt{\rightarrow\;\;\dfrac{1}{R_{eq}}\;=\;\dfrac{1\:+\:2}{20}}$

$\;\tt{\rightarrow\;\;\dfrac{1}{R_{eq}}\;=\;\dfrac{3}{20}}$

On taking reciprocal, we get

$\;\tt{\rightarrow\;\;R_{eq}\;=\;\dfrac{20}{3}}$

$\;\bf{\rightarrow\;\;\blue{R_{eq}\;=\;6.7\;\:\text{\O}mega}}$

• We see that the combination of R₁, R₂ and R₃ is in series combination with R₄ and R₅. So, equivalent resistance in series combination is given as,

$\;\sf{\rightarrow\;\;R_{eq}\;=\;R_{1,\:2\:and\:3}\;+\;R_{4}\;+\;R_{5}}$

By applying values, we get

$\;\tt{\rightarrow\;\;R_{eq}\;=\;6.7\;+\;10\;+\;10}$

$\;\bf{\rightarrow\;\;\red{R_{eq}\;=\;26.7\;\:\text{\O}mega}}$

This is the required answer.

$\;\underline{\boxed{\tt{Equivalent\;\: resistance\;\:of\;\:letter\;=\;\bf{\purple{26.7\;\:\text{\O}mega}}}}}$