Four resistances of 15 $$\Omega$$, 12 $$\Omega$$, 4 $$\Omega$$ and 10 $$\Omega$$ respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10 $$\Omega$$ to balance the network is __________ $$\Omega$$.

We have four resistances connected successively around a Wheatstone network: $$R_{AB}=15\;\Omega,\;R_{BC}=12\;\Omega,\;R_{CD}=4\;\Omega$$ and $$R_{DA}=10\;\Omega.$$ To balance a Wheatstone bridge, the ratio of the resistances in one pair of opposite arms must equal the ratio in the other pair. The condition can be written as

$$\frac{R_{BC}}{R_{AB}}=\frac{R_{CD}}{R_{DA}'}$$

where $$R_{DA}'$$ is the effective resistance of the arm DA after adding the unknown resistance $$x$$ in parallel with the given $$10\;\Omega$$ resistor.

Substituting the known values, we get

$$\frac{12}{15}=\frac{4}{R_{DA}'}.$$

Now we solve for $$R_{DA}':$$

$$12\,R_{DA}'=15\times4$$

$$12\,R_{DA}'=60$$

$$R_{DA}'=\frac{60}{12}=5\;\Omega.$$

Next we express the requirement that a resistor $$x$$ is connected in parallel with the original $$10\;\Omega$$ resistor to give this equivalent value. The formula for two resistances in parallel is first stated:

For two resistances $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance $$R_{\text{eq}}$$ is

$$R_{\text{eq}}=\frac{R_1\,R_2}{R_1+R_2}.$$

Applying the formula to the present case, with $$R_1=10\;\Omega,\;R_2=x$$ and $$R_{\text{eq}}=R_{DA}'=5\;\Omega,$$ we have

$$\frac{10\,x}{10+x}=5.$$

We now clear the denominator and solve algebraically, showing every step:

$$10x=5(10+x)$$

$$10x=50+5x$$

$$10x-5x=50$$

$$5x=50$$

$$x=\frac{50}{5}=10\;\Omega.$$

Thus, the resistance that must be connected in parallel with the existing $$10\;\Omega$$ resistor is also $$10\;\Omega.$p>

So, the answer is $$10\;\Omega.$$