Solve 10

10.  In the following circuit diagram the value of resistance X for the potential difference between B and D is zero -

a)  4 ohm
b)  6 ohm
c)  8 ohm
d)  9 ohm

 

The circuit diagram represents a Wheatstone bridge circuit. I had drawn an equivalent circuit as shown below:

Therefore when the bridge is balanced(i.e., when the potential difference between B and D is zero), the following resistance ratios are equal:

$\frac{R_2}{R_1}=\frac{R_4}{R_3}$
Now for the equivalent resistance of the set of parallel resistances $8\Omega and X\Omega$ is given by

$R_{eq}=\frac{8\times X}{(8+X)}\Omega$
Now, plugging these in the above equivalence condition, we get 



$\frac{{\displaystyle \left(\left(\frac{8\times X}{\left(8+X\right)}\Omega \right)+3\Omega \right)}}{\left(15\Omega + 6\Omega \right)}=\frac{\left(\left({\displaystyle \frac{4\times 4}{\left(4+4\right)}\Omega }\right){\displaystyle +}{\displaystyle 4}{\displaystyle \Omega }\right)}{\left(\left({\displaystyle \frac{6\times 6}{\left(6+6\right)}\Omega }\right){\displaystyle +}{\displaystyle 15}{\displaystyle \Omega }\right)}$

$$\begin{gathered} \Rightarrow \frac{\left(\left({\displaystyle \frac{8\times X}{\left(8+X\right)}\Omega }\right)+3\Omega \right)}{21\Omega }=\frac{6\Omega }{18\Omega } \\ \Rightarrow \left(\left(\frac{{\displaystyle 8\times X}}{{\displaystyle \left(8+X\right)}}\Omega \right)+3\Omega \right)=\frac{6\Omega \times 21\Omega }{18\Omega } \\ \Rightarrow \left(\frac{{\displaystyle 8\times X}}{{\displaystyle \left(8+X\right)}}\Omega \right)+3\Omega =7\Omega \\ \Rightarrow \left(\frac{{\displaystyle 8\times X}}{{\displaystyle \left(8+X\right)}}\Omega \right)=4\Omega \\ \Rightarrow 8\times X=(8+X)·4 \\ \Rightarrow 8·X=32 + 4·X \\ \Rightarrow 8·X - 4·X = 32\Omega \\ \Rightarrow 4·X=32\Omega \\ \Rightarrow X=\frac{32\Omega }{4}=8\Omega \end{gathered}$$

Therefore the correct option is (c) 8 ohm