Question

Calculate the value of x if equivalent resistance between A and B is 4 ohm.

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Answer 1

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✝️ Required Answer:

✏GiveN:

• A diagram showing the resistors in a particular arrangement.
• Equivalent resistance is 4Ω

✏To FinD:

• Find the resistor x ?

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✝️ How to solve?

Before solving this question, we need to know about the Series and parallel connection or arrangement of resistors. Series means side by side:

-------------[R1]----------[R2]-------------- In this arrangement the current remains constant but Potential difference varies.

✒To find Equivalent resistance:

$\large{\boxed{\rm{R_{eq.}=R1 + R2 + R3....}}}$

Parallel means in a parallel pattern:

----------[R1]--------

----------[R2]------- In this arrangement, the current varies, it gets divided whereas potential difference remains same.

✒To find Equivalent resistance:

$\large{\boxed{\rm{R_{eq.} = \frac{1}{R1} + \frac{1}{R2}....}}}$

By keeping these concepts in mind, let's solve this question.

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✝️ Solution:

Here,

• 4Ω and 8Ω are in series
• xΩ and 5Ω are in series
• Equivalent of 4Ω and 8Ω, xΩ and 5Ω are in parallel.

So, by using formula,

Equivalent resistance of 4Ω and 8Ω:

$\large{ \rm{ \longrightarrow \: R_{eq1} = 4 \Omega \: + 8 \Omega}} \\ \\ \large{ \rm{ \longrightarrow \: R_{eq1} = \boxed{\rm{ 12 \Omega}}}}$

Equivalent resistance of xΩ and 8Ω:

$\large{ \rm{ \longrightarrow \: \boxed{R_{eq2} = x \Omega + 8 \Omega}}}$

According to question,

$\large{ \rm{R_{eq1} \: and \: R_{eq2} \: are \: in \: parallel}}$

And Total Equivalent resistance is 4Ω

By using formula,

$\large{ \rm{ \longrightarrow \: \frac{1}{R_{Teq.}} = \frac{1}{R_{eq1}} + \frac{1}{R_{eq2}} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{1}{4} = \frac{1}{12} + \frac{1}{x + 5} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{3 - 1}{12} = \frac{1}{x + 5} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{1}{6} = \frac{1}{x + 5} }} \\ \\ \large{ \rm{ \longrightarrow \: x = \boxed{ \red{1 \Omega}}}}$

Hence, resistance of x is 1Ω.

$\large{ \therefore{ \underline{ \underline{ \green{ \rm{Hence \: solved \: \dag}}}}}}$

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