Three resistors each of resistance `1Omega` are connected in parallel. Such connection is again connected with `5/3Omega` resistor in series. The resultant resistant will be ……….`Omega`

Explanation

To find the resultant resistance of the given circuit, we will follow these steps:

### Step 1: Calculate the equivalent resistance of the three resistors in parallel.

When resistors are connected in parallel, the formula for the equivalent resistance \( R_p \) is given by:

\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

In this case, all three resistors have a resistance of \( 1 \Omega \):

\[ \frac{1}{R_p} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \]

Thus,

\[ R_p = \frac{1}{3} \Omega \]

### Step 2: Add the equivalent resistance of the parallel combination to the series resistor.

Now, we have the equivalent resistance of the parallel resistors \( R_p = \frac{1}{3} \Omega \) and a series resistor of \( \frac{5}{3} \Omega \). The total resistance \( R_t \) in series is given by:

\[ R_t = R_p + R_s \]

Where \( R_s = \frac{5}{3} \Omega \):

\[ R_t = \frac{1}{3} + \frac{5}{3} = \frac{1 + 5}{3} = \frac{6}{3} = 2 \Omega \]

### Final Result

The resultant resistance of the entire circuit is:

\[ \boxed{2 \Omega} \]

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