Two known resistances of $$R \Omega$$ and $$2R \Omega$$ and one unknown resistance $$X \Omega$$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is $$X \Omega$$, then the value of X is __________ $$\Omega$$.

From the diagram, there are two parallel paths between A and B:

• bottom branch: just R
• top branch: 2R in series with X → total = 2R + X

So equivalent resistance:

$$R_{eq}=\frac{\left(R(2R+X)\right)}{(R+2R+X)}$$

Given: Req = X

So,

$$R(2R+X)/(3R+X)=X$$

Cross-multiply:

$$R(2R+X)=X(3R+X)$$

Expand:

$$2R^2+RX=3RX+X^2$$

Bring everything to one side:

$$2R^2+RX−3RX−X^2=0$$
$$2R^2−2RX−X^2=0$$

Rearrange:

$$X^2+2RX−2R^2=0$$

Now solve quadratic in X:

$$X=[−2R\pm\sqrt{(4R^2+8R^2)}]/2$$
$$X=[−2R\pm\sqrt{\ 12R^2}]/2$$
$$X=[−2R\pm2\sqrt{3}R]/2$$

$$X=−R\pm\sqrt{3}R$$

Physical resistance must be positive, so:

$$X=R(\sqrt{3}−1)$$