Question
Find the equivalent resistance between the terminals A and B in the network shown in Figure. Given each resistor R is $10 \Omega$.
✓
Answer
Imagine a battery of emf , having no internal resistance, connected between the points A and B. The distribution of current through various branches is as shown in Figure.
Applying Kirchhoff's second law to loop KLOPK, we get
$\begin{array}{l} I _1 R +\left( I _1- I _2\right) R -2\left( I - I _1\right) R =0 \\ \text { or } 4 I _1- I _2=2 I \ldots \text { (i) }\end{array}$
Similarly, from the loop LMNOL, we have
$\begin{array}{l}2 I _2 R -\left( I - I _2\right) R -\left( I _1- I _2\right) R =0 \\ \text { or }- I _1+4 I _1= I \ldots \text { (ii) }\end{array}$
From the loop AKPONBEA, we have
$2\left( I - I _1\right) R +\left( I - I _2\right) R =\varepsilon \ldots($ (iii $)$
Solving equations (i) and (ii), we get
$I_1=\frac{3}{5} I$ and $I_2=\frac{2}{5} I$
Substituting these values in equation (iii), we get
$\begin{array}{l}2\left(I-\frac{3}{5} I\right) R+\left(I-\frac{2}{5} I\right) R=\varepsilon \\
\text { or } \frac{7}{5} I R=\varepsilon \ldots \text { (iv) }\end{array}$
If $R^{\prime}$ is the equivalent resistance between A and B , then
$I R^{\prime}=\varepsilon \ldots( v )$
From (iv) and (v), $I R^{\prime}=\frac{7}{5} I R$
or $R^{\prime}=\frac{7}{5} R=\frac{7}{5} \times 10=14 \Omega$
Applying Kirchhoff's second law to loop KLOPK, we get
$\begin{array}{l} I _1 R +\left( I _1- I _2\right) R -2\left( I - I _1\right) R =0 \\ \text { or } 4 I _1- I _2=2 I \ldots \text { (i) }\end{array}$
Similarly, from the loop LMNOL, we have
$\begin{array}{l}2 I _2 R -\left( I - I _2\right) R -\left( I _1- I _2\right) R =0 \\ \text { or }- I _1+4 I _1= I \ldots \text { (ii) }\end{array}$
From the loop AKPONBEA, we have
$2\left( I - I _1\right) R +\left( I - I _2\right) R =\varepsilon \ldots($ (iii $)$
Solving equations (i) and (ii), we get
$I_1=\frac{3}{5} I$ and $I_2=\frac{2}{5} I$
Substituting these values in equation (iii), we get
$\begin{array}{l}2\left(I-\frac{3}{5} I\right) R+\left(I-\frac{2}{5} I\right) R=\varepsilon \\
\text { or } \frac{7}{5} I R=\varepsilon \ldots \text { (iv) }\end{array}$
If $R^{\prime}$ is the equivalent resistance between A and B , then
$I R^{\prime}=\varepsilon \ldots( v )$
From (iv) and (v), $I R^{\prime}=\frac{7}{5} I R$
or $R^{\prime}=\frac{7}{5} R=\frac{7}{5} \times 10=14 \Omega$
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