Three resistors of `1 Omega, 2 Omega and 3 Omega` are connected in parallel. The combined resistance of 3 resistors should be

Options

A. greater than `3 Omega`
B. lesser than `1 Omega`
C. equal to `2 Omega`
D. between `1 Omega and 3 Omega`

Answer

To find the combined resistance of three resistors connected in parallel, we can use the formula for parallel resistances:

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

Where:
- \( R_p \) is the equivalent resistance of the parallel combination,
- \( R_1, R_2, R_3 \) are the resistances of the individual resistors.

Given:
- \( R_1 = 1 \, \Omega \)
- \( R_2 = 2 \, \Omega \)
- \( R_3 = 3 \, \Omega \)

### Step 1: Substitute the values into the formula

\[ \frac{1}{R_p} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \]

### Step 2: Calculate each term on the right side

\[ \frac{1}{1} = 1 \]
\[ \frac{1}{2} = 0.5 \]
\[ \frac{1}{3} \approx 0.333 \]

### Step 3: Add the fractions together

\[ \frac{1}{R_p} = 1 + 0.5 + 0.333 \]

To add these, we can convert them to a common denominator. The least common multiple of 1, 2, and 3 is 6.

- Convert \( 1 \) to sixths: \( 1 = \frac{6}{6} \)
- Convert \( 0.5 \) to sixths: \( 0.5 = \frac{3}{6} \)
- Convert \( 0.333 \) to sixths: \( 0.333 \approx \frac{2}{6} \)

Now we can add:

\[ \frac{1}{R_p} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{6 + 3 + 2}{6} = \frac{11}{6} \]

### Step 4: Take the reciprocal to find \( R_p \)

\[ R_p = \frac{6}{11} \, \Omega \]

### Final Answer

The combined resistance of the three resistors in parallel is:

\[ R_p \approx 0.545 \, \Omega \]

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