Question

Calculate the equivalent resistance of the following combination of resistor r₁, r₂, r₃, and r₄

Answer 1

In the given network, the series combination of resistors, r₁ and r₂ is connected in series with the parallel combination of resistors, r₃ and r₄

Equivalent resistance of resistor r₁ and r₂ , Rs= r₁ + r₂

Equivalent resistance of resistor r₃ and r₄ R_p  = `[1/ "r"^3 + 1/"r"^4]^-1 = ("r"_3"r"_4)/("r"_3 + "r"_4)`

equivalent resistance of the given network, R = R_s  + R_p = r₁ + r₂ + `("r"_3"r"_4)/("r"_3 + "r"_4)`

Answer 2

Since r₃ and r₄ are in parallel

∴ Equivalent resistance R₁ of this combination is given by

`1/"R"_1 = 1/"r"_3 + 1/"r"_4 = ("r"_4 + "r"_3)/("r"_4 "r"_3)`

or R₁ = `("r"_3 "r"_4)/("r"_3 + "r"_4)`

Now r₁, r₂ and R₁ are in series

∴ Equivalent resistance R of the whole combination is

R = r₁ + r₂ + `("r"_3 "r"_4)/("r"_3 + "r"_4)`

= `(("r"_1 + "r"_2)("r"_3 + "r"_4) + "r"_3"r"_4)/("r"_3 + "r"_4) Omega`