Math, asked by Anonymous, 2 months ago

Calculate equivalent Resistance between points P and Q.

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Answers

Answered by gotoo000612y
87

Analysis

Here we're given a circuit diagram in which a 1Ω resistor connected with two 3Ω resistors in parallel which is further connect with three 3Ω resistors in parallel. And we've to find the total effective resistance between the points P and Q. And we know that :

{\dashrightarrow{\bf{Resistors\:in\:parallel\Bigg[R_{eff}=\dfrac{R_1R_2}{R_1+R_2}\Bigg]}}}

{\dashrightarrow{\bf{Resistors\:in\:series\Big[R_{eff}=R_1+R_2\Big]}}}

Given

  • Resistor in series=1Ω
  • Resistors in parallel=Two 3Ω
  • Resistors in parallel=Three 3Ω

To Find

The total effective resistance between P and Q

Answer

_________________________

Refer to the pic provided for simplification of parallel resistors :)

_________________________

Now we've simplified all the resistors in parallel. And are left with only Three resistors in series. So let's solve them and find the equivalent resistance ahead »

\implies\rm{R_{eff}=R_1+R_2+R_2}

{\implies{\rm{R_{eff}=1 \Omega+\dfrac{3}{2} \Omega+1 \Omega}}}

\implies\rm{R_{eff}=\dfrac{3}{2} \Omega+2 \Omega}

\implies\rm{R_{eff}=\dfrac{\big(2\times2\big)+3}{2} \Omega}

\implies\rm{R_{eff}=\dfrac{4+3}{2} \Omega}

\implies\rm{R_{eff}=\dfrac{7}{2} \Omega}

\implies\rm{R_{eff}=3.5 \Omega}

{\boxed{\boxed{\bf{\therefore Equivalent\:Resistance=3.5 \Omega\checkmark}}}}

Hence the equivalent resistance between P and Q is 3.5Ω which is the required answer.

\maltese\large{\underline{\underline{\bf{Request\leadsto}}}}

View the answer in the website to see correct coding as there Ømega will be displayed as Ω.

HOPE IT HELPS.

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Answered by amitnrw
8

Given : A circuit

To Find :  equivalent Resistance between points P and Q.

Solution:

in parallel  1/R = 1/R₁  + 1/R₂ + ..  + 1/Rₙ

in series = R = R₁  + R₂ + ..  + Rₙ

3 resistance of 3 Ω  are in parallel

1/R = 1/3 + 1/3 + 1/3

=> 1/R = 3/3

=> 1/R = 1

=> R = 1 Ω

2 Resistance of 3 Ω  are in parallel

=> 1/R = 1/3 + 1/3

=> 1/R = 2/3

=> R = 3/2

=> R = 1.5  Ω

Now 1 Ω , 1.5 Ω   and 1 Ω  are in series

Hence R = 1 + 1.5 + 1  

= 3.5 Ω

equivalent Resistance between points P and Q = 3.5 Ω

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