Question

calculate the equivalent resistance for each case​

Answer 1

Answer:

(I) 15 Ω

(II) 14 Ω

Explanation:

To know :

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in series, the equivalent resistance is given by

      R = R₁ + R₂ + R₃ + ...

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in parallel, the equivalent resistance is given by

      1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...

Solution :

(I) Given resistances,

• R₁ = 10 Ω

• R₂ = 20 Ω

• R₃ = 30 Ω

R₁ and R₂ are connected in series. Their equivalent resistance, R₁₂ = R₁ + R₂

R₁₂ = 10 Ω + 20 Ω

R₁₂ = 30 Ω

Replace R₁ and R₂ by R₁₂. Now, R₁₂ and R₃ are in parallel combination.

Equivalent resistance = R₁₂₃

$\sf \dfrac{1}{R_{123}}=\dfrac{1}{R_{12}}+\dfrac{1}{R_3} \\\\ \sf \dfrac{1}{R_{123}}=\dfrac{1}{30}+\dfrac{1}{30} \\\\ \sf \dfrac{1}{R_{123}}=\dfrac{2}{30} \\\\ \sf \dfrac{1}{R_{123}}=\dfrac{1}{15} \\\\ \sf R_{123}=15 \Omega$

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(II) Given resistances,

• R₁ = 6 Ω
• R₂ = 12 Ω
• R₃ = 24 Ω

R₂ and R₃ are in parallel combination. Hence, the equivalent resistance is

$\sf \dfrac{1}{R_{23}}=\dfrac{1}{12}+\dfrac{1}{24} \\\\ \sf \dfrac{1}{R_{23}}=\dfrac{2}{24}+\dfrac{1}{24} \\\\ \sf \dfrac{1}{R_{23}}=\dfrac{3}{24} \\\\ \sf R_{23}=\dfrac{24}{3} \\\\ \sf R_{23}=8 \Omega$

Now, R₂₃ and R₁ are in series combination.

The equivalent resistance, R₁₂₃ = R₁ + R₂₃

R₁₂₃ = 6 + 8

R₁₂₃ = 14 Ω