Question

Three 8ohm resistor placed in parallel would provide a resistance that is equivalent to one what ohm resistance​

Answer 1

Answer:

$\bf 2.\bar{6}$ Ω

Explanation:

As per the provided information in the given question, we have :

• Three 8Ω resistors are placed in parallel.

We have been asked to calculate the equivalent resistance.

When the resistors are connected in parallel combination, then equivalent resistance is given by :

$\bigstar \; \boxed{ \sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \dots + \dfrac{1}{R_n} }}$

Here,

• R₁ = 8Ω
• R₂ = 8Ω
• R₃ = 8Ω

Substitute the values.

$\mapsto\sf{ \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} }$

$\mapsto\sf{ \dfrac{1}{R_p} = \Bigg \lgroup \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} \Bigg \rgroup \; \Omega }$

$\mapsto\sf{ \dfrac{1}{R_p} = \Bigg \lgroup \dfrac{1 + 1 + 1}{8} \Bigg \rgroup \; \Omega }$

$\mapsto\sf{ \dfrac{1}{R_p} = \Bigg \lgroup \dfrac{3}{8} \Bigg \rgroup \; \Omega }$

On reciprocating both sides,

$\mapsto\sf{ R_p = \Bigg \lgroup \dfrac{8}{3} \Bigg \rgroup \; \Omega }$

$\mapsto\underline{\boxed{\sf{ R_p = 2.\bar{6} \; \Omega}} }$

Therefore, the equivalent resistance is $\bf 2.\bar{6}$ Ω.

$\rule{200}2$

Learn More :

When the resistors are connected in parallel combination, then equivalent resistance is given by :

$\bigstar \; \boxed{ \sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \dots + \dfrac{1}{R_n} }}$

When the resistors are connected in series combination, then equivalent resistance is given by :

$\bigstar \; \boxed{ \sf { R_s = R_1 + R_2\dots + R_n }}$

$\rule{200}2$

Answer 2

Answer:

Given :

• Three 8 ohm resistor placed in parallel would provide a resistance that is equivalent to one.

To Find :-

• What is the equivalent resistance.

Formula Used :-

$\clubsuit$ Equivalent Resistance for parallel connection formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{\dfrac{1}{R_{eq}} =\: \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + . . . . + \dfrac{1}{R_n}}}}\: \: \: \bigstar$

where,

• $\sf R_{eq}$ = Equivalent Resistance
• R₁ = Resistance of resistor R₁
• R₂ = Resistance of resistor R₂
• R₃ = Resistance of resistor R₃

Solution :-

Given :

• ➳ R₁ = 8 Ω
• ➳ R₂ = 8 Ω
• ➳ R₃ = 8 Ω

According to the question by using the formula we get,

$\implies \sf\bold{\purple{\dfrac{1}{R_{eq}} =\: \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}}}$

$\implies \sf \dfrac{1}{R_{eq}} =\: \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}$

$\implies \sf \dfrac{1}{R_{eq}} =\: \dfrac{1 + 1 + 1}{8}$

$\implies \sf \dfrac{1}{R_{eq}} =\: \dfrac{2 + 1}{8}$

$\implies\sf \dfrac{1}{R_{eq}} =\: \dfrac{3}{8}$

By doing cross multiplication we get,

$\implies \sf 3(R_{eq}) =\: 8(1)$

$\implies \sf 3 \times R_{eq} =\: 8 \times 1$

$\implies \sf 3R_{eq} =\: 8$

$\implies \sf R_{eq} =\: \dfrac{8}{3}$

$\implies \sf\bold{\red{R_{eq} =\: 2.67\: Ω}}$

$\therefore$ The equivalent resistance is 2.67 Ω .

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EXTRA INFORMATION :-

$\clubsuit$ Equivalent Resistance for series connection formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{R_{eq} =\: R_1 + R_2 + R_3 + . . . . + R_n}}}\: \: \bigstar$

where,

• $\sf R_{eq}$ = Equivalent Resistance
• R₁ = Resistance of resistor R₁
• R₂ = Resistance of resistor R₂
• R₃ = Resistance of resistor R₃