Solved Problem on Resistors
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Find the equivalent resistance of the circuit shown in the figure. Data: R1 = 2 Ω and R2 = 4 Ω.
Problem data:
- R1 = 2 Ω;
- R2 = 4 Ω.
The circuit represents an association of infinite resistors, let's call the equivalent resistor between points A and B of X (Figure 1).
If we separate the first two resistors on the left, the association on the right side of points A' and B', highlighted in red in Figure 2, is equal to the original circuit of the problem.
The equivalent resistance to the right of A' and B' is also equal to X, and the circuit can be represented as a finite association in Figure 3.
We have two resistors in parallel (R2 and X) in series with resistor R1.
The equivalent resistance of two parallel resistors, Rpar, is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{R_{par}=\frac{R_{A}R_{B}}{R_{A}+R_{B}}}
\end{gather}
\]
with RA = R2 and RB = X. Adding this value to
resistor R1 in series, we obtain the value of the equivalent resistance X of
the problem
\[
\begin{gather}
X=R_{1}+\frac{R_{2}X}{R_{2}+X}
\end{gather}
\]
substituting the values given in the problem for R1 and R2
\[
\begin{gather}
X=2+\frac{4X}{4+X}
\end{gather}
\]
multiplying both sides of the equation by (4+X)
\[
\begin{gather}
X(4+X)=2(4+X)+\frac{4X}{4+X}(4+X)\\[5pt]
4X+X^{2}=8+2X+4X\\[5pt]
X^{2}+4X-2X-4X-8=0\\[5pt]
X^{2}-2X-8=0
\end{gather}
\]
This is a Quadratic Equation where the unknown is the value, X.
Solution of the equation \( X^{2}-2X-8=0 \)
\[
\begin{gather}
\Delta=b^{2}-4ac=(-2)^{2}-4\times 1\times(-8)=4+32=36\\[10pt]
X=\frac{-b\pm \sqrt{\Delta\;}}{2a}=\frac{-2\pm \sqrt{36\;}}{2\times 1}=\frac{2\pm 6}{2}
\end{gather}
\]
the two roots of the equation are
\[
\begin{gather}
X=4\;\Omega \qquad \text{ou} \qquad X=-2\;\Omega
\end{gather}
\]
As there is no negative resistance, the equivalent resistance is 4 Ω.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .