Solved Problem on Resistors
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Find the equivalent resistance between points A and B of the circuit shown in the figure.

Solution

Let us redraw the circuit as follows to make it easier to see (Figure 1)

Figure 1

This type of circuit is solved using the technique called Y-Δ transform (also called Star-Delta) by making the following modification to the circuit (Figure 2)

Figure 2

The resistor Ra will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{a}=\frac{R_{1}R_{2}}{R_{1}+R_{2}+R_{3}}} \end{gather} \]
\[ \begin{gather} R_{a}=\frac{2RR}{2R+R+2R}\\[5pt] R_{a}=\frac{2R^{2}}{5R}\\[5pt] R_{a}=\frac{2}{5}R \tag{I} \end{gather} \]
The resistor Rb will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{b}=\frac{R_{1}R_{3}}{R_{1}+R_{2}+R_{3}}} \end{gather} \]
\[ \begin{gather} R_{b}=\frac{2R2R}{2R+R+2R}\\[5pt] R_{b}=\frac{4R^{2}}{5R}\\[5pt] R_{b}=\frac{4}{5}R \tag{II} \end{gather} \]
The resistor Rc will be given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{c}=\frac{R_{2}R_{3}}{R_{1}+R_{2}+R_{3}}} \end{gather} \]
\[ \begin{gather} R_{c}=\frac{R2R}{2R+R+2R}\\[5pt] R_{c}=\frac{2R^{2}}{5R}\\[5pt] R_{c}=\frac{2}{5}R \tag{III} \end{gather} \]
Using the values of (I), (II), and (III), the circuit to be solved becomes the following (Figure 3)

Figure 3

The two resistors between points D and E are in series, and the equivalent resistor is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{eq}=\sum _{i=1}^{n}R_{i}} \end{gather} \]
the equivalent resistor R4 between them will be
\[ \begin{gather} R_{4}=\frac{4}{5}R+R \end{gather} \]
multiplying the numerator and denominator of the second term on the right-hand side of the equation by 5
\[ \begin{gather} R_{4}=\frac{4}{5}R+\frac{5}{5}R\\[5pt] R_{4}=\frac{9}{5}R \end{gather} \]
The two resistors between points D and F are in series, and the equivalent resistor R5 between them will be
\[ \begin{gather} R_{5}=\frac{2}{5}R+2R \end{gather} \]
multiplying the numerator and denominator of the second term on the right-hand side of the equation by 5
\[ \begin{gather} R_{5}=\frac{2}{5}R+\frac{5}{5}.2R\\[5pt] R_{5}=\frac{2}{5}R+\frac{10}{5}R\\[5pt] R_{5}=\frac{12}{5}R \end{gather} \]
The circuit is represented in (Figure 4)

Figure 4

The two resistors obtained above are connected in parallel, and the equivalent resistor is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{par}=\frac{R_{A}R_{B}}{R_{A}+R_{B}}} \end{gather} \]
the equivalent resistor R6 between them will be
\[ \begin{gather} R_{6}=\frac{R_{4}R_{5}}{R_{4}+R_{5}}\\[5pt] R_{6}=\frac{\dfrac{9}{5}R \dfrac{12}{5}R}{\dfrac{9}{5}R+\dfrac{12}{5}R}\\[5pt] R_{6}=\frac{\dfrac{108}{25}R^{2}}{\dfrac{21}{5}R}\\[5pt] R_{6}=\frac{\cancelto{36}{108}}{\cancelto{5}{25}}R^{\cancel{2}}\frac{\cancelto{1}{5}}{\cancelto{7}{21}}\frac{1}{\cancel{R}}\\[5pt] R_{6}=\frac{36}{5}\frac{1}{7}R\\[5pt] R_{6}=\frac{36}{35}R \end{gather} \]
The circuit reduces to two resistors in series (Figure 5)

Figure 5

the equivalent resistance Req of the circuit will be
\[ \begin{gather} R_{eq}=\frac{2}{5}R+\frac{36}{35}R \end{gather} \]
multiplying the numerator and denominator of the first term on the right-hand side of the equation by 7
\[ \begin{gather} R_{eq}=\frac{7}{7}.\frac{2}{5}R+\frac{36}{35}R\\[5pt] R_{eq}=\frac{14}{35}R+\frac{36}{35}R\\[5pt] R_{eq}=\frac{\cancelto{10}{50}}{\cancelto{7}{35}}R \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {R_{eq}=\frac{10}{7}R} \end{gather} \]
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