# Would it be possible to get some help with this question?. I dont know how to redraw this circuit and how to analyze parallel and seies for resistors. Thanks for any help.

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valentin68 | College Teacher | (Level 3) Associate Educator

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All cases are in the figure below. Equivalent resistances are

Between A) and B)

`R_("eq") = R1 "parallel" (R2 "series" R3) = R1 "parallel" (R2+R3) = (R1*(R2+R3))/(R1+R2+R3)`

Between A) and C)

`R_(eq) = (R1 "series" R2) "parallel" R3 = (R1+R2) "parallel" R3 =(R1+R2)*R3/(R1+R2+R3)`

Between A) and D)

`R_(eq) = ((R1 "series" R2) "parallel" R3) "series" R4 = (R1+R2) "parallel" R3) + R4 = ((R1+R2)R3)/(R1+R2+R3) + R4`

Between B) and C)

`R_(eq) = R2 "parallel" (R1 "series" R3) = R2 "parallel" (R1+R3) = R2(R1+R3)/(R1+R2+R3)`

Between B) and D)

`R_(eq) = R2 "series" R4 = R2 +R4`

Between C) and D)

`R_(eq) = R4`

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llltkl | College Teacher | (Level 3) Valedictorian

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The question has multiple sections. I am to answer only one and leaving the rest for the asker to try.

Equivalent resistance between A and D: (please see the figure for derivation of the equivalent resistance).

Remember that equivalent resistance of two resistors in series is (R1+R2) and equivalent resistance of two resistors in parallel circuit is (1/R1+1/R2).

Step 1. Between A and D, R1 and R2 are i series, so eqv. resistor=(R1+R2)

Step 2. (R1+R2) and R3 are in parallel, so

`1/(eqv. resistance)=1/(R1+R2)+1/(R3)`

`Eqv. resistance=(R3(R1+R2))/(R1+R2+R3)`

Step 3. `(R3(R1+R2))/(R1+R2+R3)` and R4 are in series, so

Eqv. resistor`=(R4+(R3(R1+R2))/(R1+R2+R3))`

This is the equivalent resistance between A and D.

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