If two resistors of `25 Omega` and `15 Omega` are joined together in series and then placed in parallel with a `40 Omega` resistor, the effective resistance of the combination is:

Options

A. `0.1 Omega`
B. `10Omega`
C. `20Omega`
D. `40Omega`

Answer

To find the effective resistance of the given combination of resistors, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistors**: We have two resistors, \( R_1 = 25 \, \Omega \) and \( R_2 = 15 \, \Omega \), which are connected in series.

2. **Calculate the Series Resistance**: The total resistance for resistors in series is simply the sum of their resistances:
\[ R_s = R_1 + R_2 = 25 \, \Omega + 15 \, \Omega = 40 \, \Omega \]

3. **Identify the Parallel Resistor**: Now, this series combination (which is \( 40 \, \Omega \)) is placed in parallel with another resistor \( R_3 = 40 \, \Omega \).

4. **Calculate the Effective Resistance for Parallel Resistors**: The formula for the equivalent resistance \( R_{eq} \) of two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{R_3} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{40 \, \Omega} + \frac{1}{40 \, \Omega} \]

5. **Combine the Fractions**:
\[ \frac{1}{R_{eq}} = \frac{1 + 1}{40} = \frac{2}{40} = \frac{1}{20} \]

6. **Calculate \( R_{eq} \)**: Taking the reciprocal gives us:
\[ R_{eq} = 20 \, \Omega \]

### Final Answer:
The effective resistance of the combination is \( 20 \, \Omega \).

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