Find equivalent resistance across a nad b
Answers
Answer:
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Circuit analysis:
This circuit can easily be solved by applying mirror symmetry.
Please refer to the attachment first !!
Draw an imaginary line passing through the center of the circuit. You will notice that the RHS is the exact mirror image of the LHS.
As both the sides are symmetrical no current will be divided at point O (as shown in the attachment).
Now rearrange the circuit and solve.
Between point C and D:
R₁and R₂ are connected in series with each other,
⇒ R₁₂ = R₁ + R₂
⇒ R₁₂ = 9 + 9
⇒ R₁₂ = 18 kΩ
Now,
R₁₂₃ is connected in parallel to R₃,
⇒ 1/R₁₂₃ = 1/R₁₂ + 1 / R₃
⇒ 1/R₁₂₃ = 1/18 + 1/9
⇒ 1/R₁₂₃ = 1+2/18
⇒ 1/R₁₂₃ = 3/18
⇒ R₁₂₃ = 6kΩ
R₄ and R₅ are connected in series to each other (refer to the attachment)
⇒ R₄₅ = R₄ + R₅
⇒ R₄₅ = 9 +9
⇒ R₄₅ = 18 kΩ
R₁₂₃ and R₄₅ are connected in parallel.
⇒ 1 / R₁₂₃₄₅ = 1 / R₁₂₃ + 1/ R₄₅
⇒ 1 / R₁₂₃₄₅ = 1 / 6 + 1 / 18
⇒ 1 / R₁₂₃₄₅ = 1 + 3 / 18
⇒ 1 / R₁₂₃₄₅ = 4 / 18
⇒ R₁₂₃₄₅ = 4.5 kΩ
Between a and b :
R₆, R₇, and R₁₂₃₄₅ are connected series,
⇒ R₁₂₃₄₅₆₇ = R₁₂₃₄₅ + R₆ + R₇
⇒ R₁₂₃₄₅₆₇ = 4.5+ 9 + 9
⇒ R₁₂₃₄₅₆₇ = 22.5 kΩ
R₈ and R₉ are connected series,
⇒ R₈₉ = R₈ + R₉
⇒ R₈₉ = 9 + 9
⇒ R₈₉ =18 kΩ
R₁₂₃₄₅₆₇ is connected in parallel with R₈₉,
⇒ 1 /R = 1/R ₁₂₃₄₅₆₇ + 1 / R₈₉
⇒ 1 /R = 1 /22.5 + 1 / 18
⇒ 1 /R = 18 + 22.5 / 18 x 22.5
⇒ R = 405 / 40.5
⇒ R = 10 kΩ
The equivalent resistance of the circuit is 10kΩ.
