Question
A $$12\,cm$$ wire is given a shape of a right angled triangle $$ABC$$ having sides $$3\,cm,4\,cm$$ and $$5\,cm$$ as shown in the figure. The resistance between two ends $$\left( {AB,BC,CA} \right)$$ of the respective sides are measured one by one by a multi-meter. The
resistances will be in the ratio of
A $$12\,cm$$ wire is given a shape of a right angled triangle $$ABC$$ having sides $$3\,cm,4\,cm$$ and $$5\,cm$$ as shown in the figure. The resistance between two ends $$\left( {AB,BC,CA} \right)$$ of the respective sides are measured one by one by a multi-meter. The
resistances will be in the ratio of
A.
$$3:4:5$$
B.
$$9:16:25$$
C.
$$27:32:35$$
D.
$$21:24:25$$
Answer :
$$27:32:35$$
Solution :
Resistance is directly proportional to length
$$\eqalign{ & \frac{1}{{{R_{AB}}}} = \frac{1}{3} + \frac{1}{{4 + 5}} = \frac{{\left( {4 + 5} \right) + 3}}{{\left( 3 \right)\left( {4 + 5} \right)}} \cr & {R_{AB}} = \frac{{3 \times \left( {4 + 5} \right)}}{{3 + \left( {4 + 5} \right)}} = \frac{{27}}{{12}} \cr} $$
Similarly,
$$\eqalign{ & {R_{BC}} = \frac{{4 \times \left( {3 + 5} \right)}}{{4 + \left( {3 + 5} \right)}} = \frac{{32}}{{12}} \cr & {R_{AC}} = \frac{{5 \times \left( {3 + 4} \right)}}{{5 + \left( {3 + 4} \right)}} = \frac{{32}}{{12}} \cr & \therefore {R_{AB}}:{R_{BC}}:{R_{AC}} = 27:32:35 \cr} $$
Resistance is directly proportional to length
$$\eqalign{ & \frac{1}{{{R_{AB}}}} = \frac{1}{3} + \frac{1}{{4 + 5}} = \frac{{\left( {4 + 5} \right) + 3}}{{\left( 3 \right)\left( {4 + 5} \right)}} \cr & {R_{AB}} = \frac{{3 \times \left( {4 + 5} \right)}}{{3 + \left( {4 + 5} \right)}} = \frac{{27}}{{12}} \cr} $$
Similarly,
$$\eqalign{ & {R_{BC}} = \frac{{4 \times \left( {3 + 5} \right)}}{{4 + \left( {3 + 5} \right)}} = \frac{{32}}{{12}} \cr & {R_{AC}} = \frac{{5 \times \left( {3 + 4} \right)}}{{5 + \left( {3 + 4} \right)}} = \frac{{32}}{{12}} \cr & \therefore {R_{AB}}:{R_{BC}}:{R_{AC}} = 27:32:35 \cr} $$
