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 $\text{KLOPK},$ we get
$ I _1 R +\left( I _1- I _2\right) R -2\left( I - I _1\right) R =0$
$\text { or } 4 I _1- I _2=2 I \ldots \text { (i) }$
Similarly, from the loop $\text{LMNOL},$ we have
$2 I _2 R -\left( I - I _2\right) R -\left( I _1- I _2\right) R =0$
$\text { or }- I _1+4 I _1= I \ldots \text { (ii) }$
From the loop $\text{AKPONBEA},$ 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
$2\left(I-\frac{3}{5} I\right) R+\left(I-\frac{2}{5} I\right) R=\varepsilon$
$\text { or } \frac{7}{5} I R=\varepsilon \ldots \text { (iv) }$
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 $\text{KLOPK},$ we get
$ I _1 R +\left( I _1- I _2\right) R -2\left( I - I _1\right) R =0$
$\text { or } 4 I _1- I _2=2 I \ldots \text { (i) }$
Similarly, from the loop $\text{LMNOL},$ we have
$2 I _2 R -\left( I - I _2\right) R -\left( I _1- I _2\right) R =0$
$\text { or }- I _1+4 I _1= I \ldots \text { (ii) }$
From the loop $\text{AKPONBEA},$ 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
$2\left(I-\frac{3}{5} I\right) R+\left(I-\frac{2}{5} I\right) R=\varepsilon$
$\text { or } \frac{7}{5} I R=\varepsilon \ldots \text { (iv) }$
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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