In the question stem, we're told that a circuit containing three identical resistors connected in parallel has an equivalent resistance equal to \(\displaystyle 2\:\Omega\). We are then asked to solve for the resistance of each individual resistor.

To start with, it's important to remember that resistors in parallel add inversely. Thus, the inverse of the equivalent resistance is equal to the sum of the inverse of each individual resistor. Put another way:

\(\displaystyle \frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+....\frac{1}{R_{n}}\)

Since we know the three resistors we're dealing with are identical, we can assign each of them a value of \(\displaystyle x\).

\(\displaystyle \frac{1}{2\: \Omega }=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}\)

And rearranging, we obtain:

\(\displaystyle x=(2\: \Omega)(3)=6\: \Omega\)

The equivalent resistance of resistors connected in parallel is given by the following equation,

\(\displaystyle \frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...\)

In our problem,

\(\displaystyle \frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}=\frac{2}{30}+\frac{3}{30}\)

\(\displaystyle \frac{1}{R_{p}}=\frac{5}{30}\)

\(\displaystyle R_{p}=\frac{30}{5}=6\Omega\)

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

\(\displaystyle R_{s}=\sum R_{n}=R_{1}+R_{2}+...\)

In our problem,

\(\displaystyle R_{s}=R_{1}+R_{2}=15+10=25\Omega\)

The equivalent resistance of resistors connected in series is the sum of the resistance values of each resistor, or

\(\displaystyle R_{s}=\sum R_{n}=R_{1}+R_{2}+...\)

In our problem,

\(\displaystyle R_{s}=R_{1}+R_{2}+R_{5}=15+10+5=30\Omega\)

The equivalent resistance of resistors connected in parallel is given by the following equation,

\(\displaystyle \frac{1}{R_{p}}=\sum \left ( \frac{1}{R_{n}} \right )=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+...\)

In our problem,

\(\displaystyle \frac{1}{R_{p}}=\frac{1}{15}+\frac{1}{10}+\frac{1}{5}=\frac{2}{30}+\frac{3}{30}+\frac{6}{30}\)

\(\displaystyle \frac{1}{R_{p}}=\frac{11}{30}\)

\(\displaystyle R_{p}=\frac{30}{11}=2.73\Omega\)

Remember the rules of adding resistors. For \(\displaystyle n\) resistors in series,

\(\displaystyle R_{eq}=R_{1}+R_{2}+ ... +R_{n}\)

and for \(\displaystyle n\) resistors in parallel,

\(\displaystyle R_{eq}=\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}}+...+\frac{1}{R_{n}}\right )^{-1}\)

The idea is to start from the side furthest away from the battery and work back toward it. Notice that \(\displaystyle R_{2}\) and \(\displaystyle R_{3}\) are in parallel. We can add them in parallel so that they have an equivalent resistance \(\displaystyle R_{23}\),

This can be calculated, but for 3 resistors we can leave it in equation form until the end. This looks like:

\(\displaystyle R_{23}=\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=9.375\Omega\)

Notice that \(\displaystyle R_{1}\) is now in series with \(\displaystyle R_{23}\),

\(\displaystyle R_{eq}=R_{1}+R_{23}=R_{1}+\left ( \frac{1}{15\Omega}+\frac{1}{25\Omega}\right )^{-1}=19.375\Omega\)

Since the resistors are in parallel to a \(\displaystyle 10V\) battery, the voltage drop across each resistor has to be equal to the voltage gain across the battery. Therefore, the voltage drop for any of the resistors will be \(\displaystyle 10V\).

In order to find the equivalent resistance of resistors in parallel, we add the inverses of their values, as shown below

\(\displaystyle \frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{5 \Omega}+\frac{1}{10 \Omega}+\frac{1}{1\Omega}+\frac{1}{6 \Omega}=1.467 \Omega ^{-1}\)

Finally 

\(\displaystyle \frac{1}{R_{tot}}=1.467 \Omega ^{-1} \Rightarrow R_{tot}=\frac{1}{1.467 \Omega ^{-1}}=0.68 \Omega\)

We can answer this by looking at how the equivalent resistance of a parallel circuit is calculated.

\(\displaystyle \frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+...\Rightarrow R_{tot}= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+...}\).

What we can see is that because the inverses are being added, any additional resistors will make contributions to the denominator of the fraction. Thus, lowering, or decreasing the equivalent resistance of the circuit.

The equivalent resistance of resistors in series is calculated by \(\displaystyle R_{tot}= R_1 + R_2 + R_3\), while for parallel resistors, \(\displaystyle R_{tot}= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}\). This clearly shows that resistors in a series configuration will generate a much higher, or maximum, equivalent resistance value.