Question

Two resistors are joined in parallel whose resultant is 6/5 ohm. One of the resistence wire is broken and the effective resistence becomes 2 ohm. Then the resistence in ohm of the wire that got broken was ?

Answer 1

It's easy.
If one resistor is of 2ohm then the other would be 1\x
Now,
Since they are connected in parallel,
1/2 + 1/X = 5/6
1\x = 5\6 - 1/2
1\x = 1/3
Therefore X = 3ohm
So,the other resistor was of 3ohm

Answer 2

Resistance of the wire in ohm which was broken is 3 ohm when effective resistance is 2 ohm

Solution:

The resistance of the wire in S. I. unit be calculated on the wire breakage be calculated as follows,

As given in the question that the connection was done in parallel as the resultant resistance, as $\frac{6}{5}$ ohm and one resistance be 2 ohm.

$\begin{array}{l}{\frac{1}{\text { Effective Resistance }}=\frac{1}{R_{1}}+\frac{1}{R_{2}}} \\\\ {\frac{5}{6}=\frac{1}{2}+\frac{1}{x}} \\ \\{\frac{1}{x}=\frac{5}{6}-\frac{1}{2}}\end{array}$

$\begin{array}{l}{\frac{1}{x}=\frac{5-3}{6}} \\ \\{\frac{1}{x}=\frac{1}{3}}\end{array}$

x = 3 ohm