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Question
Calculate the equivalent resistance of the following combination of resistor r1, r2, r3, and r4
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Solution 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 Rp = `[1/ "r"^3 + 1/"r"^4]^-1 = ("r"_3"r"_4)/("r"_3 + "r"_4)`
equivalent resistance of the given network, R = Rs + Rp = r1 + r2 + `("r"_3"r"_4)/("r"_3 + "r"_4)`
Solution 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`
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