Six resistances each of value r = 6,Omega are | vedclass

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(C) Let the resistance of each resistor be $r$. The circuit consists of three branches between nodes $A, B,$ and $C$: $1$. Branch $AB$ has one resisto

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$5 : 4 : 3$

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(C) Let the resistance of each resistor be $r$. The circuit consists of three branches between nodes $A, B,$ and $C$: $1$. Branch $AB$ has one resistor $r$. $2$. Branch $BC$ has two resistors $r$ in parallel,equivalent to $r/2$. $3$. Branch $AC$ has three resistors $r$ in parallel,equivalent to $r/3$. To find the equivalent resistance $R_1$ between $A$ and $B$: The branch $AB$ (resistance $r$) is in parallel with the series combination of branches $BC$ and $AC$ (resistance $r/2 + r/3 = 5r/6$). $R_1 = \frac{r \cdot (5r/6)}{r + 5r/6} = \frac{5r^2/6}{11r/6} = \frac{5}{11}r$. To find the equivalent resistance $R_2$ between $B$ and $C$: The branch $BC$ (resistance $r/2$) is in parallel with the series combination of branches $AB$ and $AC$ (resistance $r + r/3 = 4r/3$). $R_2 = \frac{(r/2) \cdot (4r/3)}{r/2 + 4r/3} = \frac{4r^2/6}{11r/6} = \frac{4}{11}r$. To find the equivalent resistance $R_3$ between $A$ and $C$: The branch $AC$ (resistance $r/3$) is in parallel with the series combination of branches $AB$ and $BC$ (resistance $r + r/2 = 3r/2$). $R_3 = \frac{(r/3) \cdot (3r/2)}{r/3 + 3r/2} = \frac{r^2/2}{11r/6} = \frac{3}{11}r$. Thus,$R_1 : R_2 : R_3 = \frac{5}{11}r : \frac{4}{11}r : \frac{3}{11}r = 5 : 4 : 3$.

Six resistances each of value $r = 6\,\Omega$ are connected between points $A, B$ and $C$ as shown in the figure. If $R_1, R_2$ and $R_3$ are the net resistance between $A$ and $B$,between $B$ and $C$,and between $A$ and $C$ respectively,then $R_1 : R_2 : R_3$ will be equal to:

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