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How To Find The Equivalent Resistance of a Complex Circuit

How To Find The Equivalent Resistance of a Complex Circuit :- To find the equivalent resistance of complex circuits we have to identify the equipotential points in the circuit and connect them. The identity of equipotential points is that the circuit is symmetrical relative to these points. In a given circuit below, there may be two symmetric axis :-

(a) Normal Symmetric Axis : This axis (YY’) is in the perpendicular direction of the flow of electric current. The electric potentials of the points situated on this axis are equal. In the above figure :-

V₆ = V₅

V₂ = V₀ = V₄

V₇ = V₈

(b) Parallel Symmetric Axis : This axis (XX’) is in the direction of flow of electric current. The electric potentials of the points situated on this axis are never equal. Electric potential decreases on this axis when moving in the direction of electric current. Hence in the above figure :-

V₁ > (V₆ = V₅) > (V₂ = V₀ = V₄) > (V₇ = V₈) > V₃

If the circuit is bent with respect to the parallel symmetric axis, then electric potentials of the overlapping points [(5 and 6), (2, 0 and 4) and (7 and 8)] being equal, the circuit can be made as follows :-

Further solving :-

Note :- We can also consider the above circuit as divided into two parts with respect to the parallel symmetry axis (XX’) and the resistance of each part will be R’ = 3R, as shown in the figure below :-

Now if we find the equivalent resistance of both the parts (both parts will be considered in parallel combination), then :-

Finding Equivalent Resistance of an Unbalanced Wheatstone Bridge

(How To Find The Equivalent Resistance of a Complex Circuit)

If in the above circuit $\displaystyle \frac{P}{R}\ne \frac{Q}{S}$ , then it becomes an unbalanced Wheatstone bridge. To solve this circuit, star-delta transformation has been done as follows :-

$\displaystyle {{R}_{YQ}}=Y+Q=\frac{PG}{P+R+G}+\frac{Q}{1}=\frac{PG+Q\left( P+R+G \right)}{\left( P+R+G \right)}$

$\displaystyle {{R}_{ZS}}=Z+S=\frac{RG}{P+R+G}+\frac{S}{1}=\frac{RG+S\left( P+R+G \right)}{\left( P+R+G \right)}$

$\displaystyle {{R}_{AB}}=X+\frac{\left( {{R}_{YQ}}\times {{R}_{ZS}} \right)}{\left( {{R}_{YQ}}+{{R}_{ZS}} \right)}$

$\displaystyle {{R}_{AB}}=X+\frac{\left[ \frac{PG+Q\left( P+R+G \right)}{\left( P+R+G \right)} \right]\times \left[ \frac{RG+S\left( P+R+G \right)}{\left( P+R+G \right)} \right]}{\left[ \frac{PG+Q\left( P+R+G \right)}{\left( P+R+G \right)} \right]+\left[ \frac{RG+S\left( P+R+G \right)}{\left( P+R+G \right)} \right]}$

$\displaystyle {{R}_{AB}}=X+\frac{\frac{\left[ PG+Q\left( P+R+G \right) \right]\times \left[ RG+S\left( P+R+G \right) \right]}{{{\left( P+R+G \right)}^{2}}}}{\frac{\left[ PG+Q\left( P+R+G \right) \right]+\left[ RG+S\left( P+R+G \right) \right]}{\left( P+R+G \right)}}$

$\displaystyle {{R}_{AB}}=X+\frac{\left[ PG+Q\left( P+R+G \right) \right]\times \left[ RG+S\left( P+R+G \right) \right]}{\left( P+R+G \right)\left\{ \left[ PG+Q\left( P+R+G \right) \right]+\left[ RG+S\left( P+R+G \right) \right] \right\}}$

$\displaystyle {{R}_{AB}}=\frac{PR}{P+R+G}+\frac{\left[ PG+Q\left( P+R+G \right) \right]\times \left[ RG+S\left( P+R+G \right) \right]}{\left( P+R+G \right)\left\{ \left[ PG+Q\left( P+R+G \right) \right]+\left[ RG+S\left( P+R+G \right) \right] \right\}}$

$\displaystyle {{R}_{AB}}=\frac{PR\left\{ \left[ PG+Q\left( P+R+G \right) \right]+\left[ RG+S\left( P+R+G \right) \right] \right\}+\left[ PG+Q\left( P+R+G \right) \right]\times \left[ RG+S\left( P+R+G \right) \right]}{\left( P+R+G \right)\left\{ \left[ PG+Q\left( P+R+G \right) \right]+\left[ RG+S\left( P+R+G \right) \right] \right\}}$

$\displaystyle {{R}_{AB}}=\frac{{{P}^{2}}RG+PRQ\left( P+R+G \right)+P{{R}^{2}}G+PRS\left( P+R+G \right)+PR{{G}^{2}}+PGS\left( P+R+G \right)+RGQ\left( P+R+G \right)+SQ{{\left( P+R+G \right)}^{2}}}{\left( P+R+G \right)\left[ (P+R)G+Q(P+R)+GQ+S(P+R)+GS \right]}$

$\displaystyle {{R}_{AB}}=\frac{\left\{ {{P}^{2}}RG+P{{R}^{2}}G+PR{{G}^{2}} \right\}+PRQ\left( P+R+G \right)+PRS\left( P+R+G \right)+PGS\left( P+R+G \right)+RGQ\left( P+R+G \right)+SQ{{\left( P+R+G \right)}^{2}}}{\left( P+R+G \right)\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{PRG\left( P+R+G \right)+PRQ\left( P+R+G \right)+PRS\left( P+R+G \right)+PGS\left( P+R+G \right)+RGQ\left( P+R+G \right)+SQ{{\left( P+R+G \right)}^{2}}}{\left( P+R+G \right)\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{PRG+PRQ+PRS+PGS+RGQ+SQ\left( P+R+G \right)}{\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{PRG+PRQ+PRS+PGS+RGQ+SQP+SQR+SQG}{\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{G\left( PR+PS+RQ+SQ \right)+PRQ+PRS+SQP+SQR}{\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{G\left[ P\left( R+S \right)+Q\left( R+S \right) \right]+PQ\left( R+S \right)+RS\left( P+Q \right)}{\left[ G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right) \right]}$

$\displaystyle {{R}_{AB}}=\frac{PQ\left( R+S \right)+RS\left( P+Q \right)+G\left( P+Q \right)\left( R+S \right)}{G\left( P+Q+R+S \right)+\left( P+R \right)\left( Q+S \right)}$

Cube Resistance Problem

(How To Find The Equivalent Resistance of a Complex Circuit)

For a cube having equal resistance along its edges, the equivalent resistance can be calculated between :-

Two vertices along the body diagonal
Two vertices along the face diagonal
Two adjacent vertices

Equivalent resistance between two vertices along the body diagonal (AG) = 5R/6

Equivalent resistance between two vertices along the face diagonal (AC) = 3R/4

Equivalent resistance between two adjacent vertices (AB) = 7R/12

For the derivation of the above results, read the article “Cube Resistance Problem | JEE Main | JEE Advanced“.

Equivalent Resistance of Circuits with Infinite Resistances

(How To Find The Equivalent Resistance of a Complex Circuit)

In the above circuit, resistances R₁ , R₂ and R₃ are repeated, hence the above circuit can be simplified as follows :-

The above circuit, when viewed from points X and Y, appears to be the same as when viewed from points A and B, hence the resistance between X and Y will be the same as the resistance between the points A and B.

Now here R_AB and R₃ are in parallel and their resultant is connected in series with R₁ and R₂, hence the total resistance between points A and B can be determined as follows :-

$\displaystyle {{R}_{AB}}={{R}_{1}}+\frac{{{R}_{3}}\times {{R}_{AB}}}{{{R}_{3}}+{{R}_{AB}}}+{{R}_{2}}$

$\displaystyle \Rightarrow {{R}_{AB}}=\frac{{{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{AB}}+{{R}_{3}}{{R}_{AB}}+{{R}_{2}}{{R}_{3}}+{{R}_{2}}{{R}_{AB}}}{\left( {{R}_{3}}+{{R}_{AB}} \right)}$

$\displaystyle \Rightarrow {{R}_{3}}{{R}_{AB}}+R_{AB}^{2}={{R}_{1}}{{R}_{3}}+{{R}_{1}}{{R}_{AB}}+{{R}_{3}}{{R}_{AB}}+{{R}_{2}}{{R}_{3}}+{{R}_{2}}{{R}_{AB}}$

$\displaystyle \Rightarrow R_{AB}^{2}-\left( {{R}_{1}}+{{R}_{2}} \right){{R}_{AB}}-\left( {{R}_{1}}+{{R}_{2}} \right){{R}_{3}}=0$

$\displaystyle {{R}_{AB}}=\frac{\left( {{R}_{1}}+{{R}_{2}} \right)+\sqrt{{{\left( {{R}_{1}}+{{R}_{2}} \right)}^{2}}+4{{R}_{3}}\left( {{R}_{1}}+{{R}_{2}} \right)}}{2}$

Similarly, if the circuit is given as follows then :-

On solving the circuit again in the same way :-

Similarly, here R_AB and R₂ are in parallel and their resultant is connected in series with R₁, hence the total resistance between points A and B can be determined as follows :-

$\displaystyle {{R}_{AB}}={{R}_{1}}+\frac{{{R}_{\text{2}}}\times {{R}_{AB}}}{{{R}_{2}}+{{R}_{AB}}}$