A `120 V` house circuit has the following light bulbs switched on : `40 W, 60 W` and `75 W`. Find the equivalent resistance of these bulbs.

Explanation

To find the equivalent resistance of the light bulbs connected in a 120 V house circuit, we will follow these steps:

### Step 1: Calculate the resistance of each bulb
The resistance of each bulb can be calculated using the formula:
\[ R = \frac{V^2}{P} \]
where \( V \) is the voltage across the bulb and \( P \) is the power rating of the bulb.

#### For the 40 W bulb:
\[ R_1 = \frac{120^2}{40} = \frac{14400}{40} = 360 \, \Omega \]

#### For the 60 W bulb:
\[ R_2 = \frac{120^2}{60} = \frac{14400}{60} = 240 \, \Omega \]

#### For the 75 W bulb:
\[ R_3 = \frac{120^2}{75} = \frac{14400}{75} = 192 \, \Omega \]

### Step 2: Use the formula for equivalent resistance in parallel
The equivalent resistance \( R_{eq} \) for resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

Substituting the values we calculated:
\[ \frac{1}{R_{eq}} = \frac{1}{360} + \frac{1}{240} + \frac{1}{192} \]

### Step 3: Calculate the individual fractions
To add these fractions, we will first find a common denominator. The least common multiple of 360, 240, and 192 is 4320.

Calculating each term:
\[ \frac{1}{360} = \frac{12}{4320} \]
\[ \frac{1}{240} = \frac{18}{4320} \]
\[ \frac{1}{192} = \frac{22.5}{4320} \]

Now, adding these fractions:
\[ \frac{1}{R_{eq}} = \frac{12 + 18 + 22.5}{4320} = \frac{52.5}{4320} \]

### Step 4: Find \( R_{eq} \)
Now, we can find \( R_{eq} \) by taking the reciprocal:
\[ R_{eq} = \frac{4320}{52.5} \approx 82.29 \, \Omega \]

### Final Answer
The equivalent resistance of the bulbs is approximately \( 82.29 \, \Omega \).

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