Two bulbs are marked 220 V, 100 W and 220 V, 50W respectively. They are connected in series to 220 V mains. Find the ratio of heats generated in them.

Answer

To solve the problem of finding the ratio of heats generated in two bulbs connected in series, we will follow these steps:

### Step 1: Understand the relationship between power, voltage, and resistance.
The power \( P \) of an electrical device is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the device and \( R \) is its resistance. Rearranging this formula gives us:
\[ R = \frac{V^2}{P} \]

### Step 2: Calculate the resistance of each bulb.
For the first bulb (220 V, 100 W):
\[ R_1 = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega \]

For the second bulb (220 V, 50 W):
\[ R_2 = \frac{(220)^2}{50} = \frac{48400}{50} = 968 \, \Omega \]

### Step 3: Use the formula for heat generated in resistors in series.
When two resistors are connected in series, the heat generated in each resistor can be expressed as:
\[ H \propto R \]
Since the current \( I \) is the same through both bulbs, the heat generated in each bulb is directly proportional to its resistance.

### Step 4: Find the ratio of heats generated in the bulbs.
The ratio of heats generated in the two bulbs can be expressed as:
\[ \frac{H_1}{H_2} = \frac{R_1}{R_2} \]
Substituting the values we calculated:
\[ \frac{H_1}{H_2} = \frac{R_1}{R_2} = \frac{484}{968} \]

### Step 5: Simplify the ratio.
\[ \frac{H_1}{H_2} = \frac{484 \div 484}{968 \div 484} = \frac{1}{2} \]

### Final Answer:
The ratio of heats generated in the two bulbs is:
\[ \text{Ratio} = 1:2 \]