Question

C) The resistors 5ohm and 6ohm resistors are connected in parallel. This combination is than
connected in series with 5ohm resistor. Calculate the total resistance.


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Answer 1

Answer :

The required total resistance is 85/11 Ω

Step-by-step explanation :

Given :

• The resistors 5 Ω and 6 Ω resistors are connected in parallel.
• This combination is then  connected in series with 5 Ω resistor.

To find :

the total resistance

Solution :

To know :

• When resistors of resistances R₁ , R₂ , .... Rₙ are connected in parallel, the equivalent resistance is given as 1/Rₚ = 1/R₁ + 1/R₂ + .... + 1/Rₙ

• When resistors of resistances R₁ , R₂ , .... Rₙ are connected in series, the equivalent resistance is given as Rₛ = R₁ + R₂ + .... + Rₙ

Let

• R₁ = 5 Ω
• R₂ = 6 Ω
• R₃ = 5 Ω

R₁ and R₂ are connected in parallel.

The equivalent resistance of the resistors R₁ and R₂ is

   $\sf \dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2} \\\\ \sf \dfrac{1}{R_{12}}=\dfrac{1}{5}+\dfrac{1}{6} \\\\ \sf \dfrac{1}{R_{12}}=\dfrac{6+5}{30} \\\\ \sf \dfrac{1}{R_{12}}=\dfrac{11}{30} \\\\ \sf R_{12}=\dfrac{30}{11} \Omega$

R₁₂ and R₃ are connected in series.

 R₁₂₃ = R₁₂ + R₃

 R₁₂₃ = 30/11 + 5

 $\sf R_{123}=\dfrac{30+55}{11} \\\\ \sf R_{123}=\dfrac{85}{11} \Omega$

The total resistance is 85/11 Ω