Question

Find the equivalent resistance of the combination of resistors shown in the figure below. (R1 = 3.32 µΩ, R2 = 11.6 µΩ.)

Answer 1

a. The equivalent resistance in series is 14.92 µΩ

b. The equivalent resistance in parallel is 2.58 µΩ

Equivalent resistance

There are two types of equivalent resistance

• equivalent resistance in series
• equivalent resistance in parallel

a. Equivalent resistance in series

The equivalent resistance in series is 14.92 µΩ

For a series circuit of two resistors R₁ and R₂, the equivalent resistance in series is R = R₁ + R₂

Since R₁ = 3.32 µΩ and R₂ = 11.6 µΩ

So, R = R₁ + R₂

R = 3.32 µΩ + 11.6 µΩ

R = 14.92 µΩ

So, the equivalent resistance in series is 14.92 µΩ

b. Equivalent resistance in parallel

The equivalent resistance in parallel is 2.58 µΩ

For a series circuit of two resistors R₁ and R₂, the equivalent resistance in parallel is thus 1/R = 1/R₁ + 1/R₂.

Thus R = R₁R₂/(R₁ + R₂)

Since R₁ = 3.32 µΩ and R₂ = 11.6 µΩ

So, R = R₁R₂/(R₁ + R₂)

R = 3.32 µΩ × 11.6 µΩ/(3.32 µΩ + 11.6 µΩ)

R = 38.512µΩ²/14.92 µΩ

R = 2.58µΩ

So, the equivalent resistance in parallel is 2.58 µΩ

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