The resistance of two conductors in parallel is $ 12 \Omega $ and in series is $ 50 \Omega $. Find the resistance of each conductor.

Give: 

Resistance of two conductor-

In parallel = $ 12 \Omega $

In series = $ 50 \Omega $

To find: The resistance of each conductor.

Solution:

Let $R_1$ and $R_2$ are the resistance of the two conductors.

Now, according to the question,

When the resistance of two conductors are connected in series, then the total resistance is given as-

$R_T=R_1+R_2$

$50=R_1+R_2$    ----------------(i)

 When the resistance of two conductors are connected in parallel, then the total resistance is given as-

$\frac {1}{R_T}=\frac {1}{R_1}+\frac {1}{R_2}$

$\frac {1}{12}=\frac {1}{R_1}+\frac {1}{R_2}$

$\frac {1}{12}=\frac {R_2+R_1}{R_1\times R_2}$      

$12=\frac {R_1\times R_2}{R_2+R_1}$         ---------------(ii) 

Putting the value of $R_1+R_2$ from (i) in the above equation we get-

$\frac {R_1\times R_2}{50}=12$ 

$R_1\times {R_2}=600$          ---------------(iii)

Substituting the value of $R_2$ from (i) in the equation (iii) we got a quadratic equation:

$R_1\times ({50-R_1})=600$     $[from\ (i)\ we\ can\ obtain\ (R_2=50-R_1)]$

$50R_1-R_1^2=600$

$R_1^2-50R_1+600=0$

$R_1^2-20R_1-30R_1+600=0$

$R_1(R_1-20)-30(R_1-20)=0$

$(R_1-30)(R_1-20)=0$

$(R_1-30)=0\Rightarrow R_1=30$

$(R_1-20)=0\Rightarrow R_1=20$

We get 2 values of $R 1$, one is 30, and the other is 20.

Now, putting the value of $R_1=30$ in (i) we get-

$50=30+R_2$  

$R_2=50-30$  

$R_2=20$  

If we put the value of $R_1=20$ in (i) we get-

$R_2=30$  

Thus, we can see that both $R_1$ and $R_2$ can have the resistance $ 20 \Omega $ or $ 30 \Omega $ interchangeably.

So, the resistance of one conductor is $ 20 \Omega $ and the resistance of the other conductor is $ 30 \Omega $.

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Updated on: 10-Oct-2022

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