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Step-by-Step Solution
Step 1: Identify the Resistors of Each Bulb
Each bulb can be treated as a resistor, where its resistance is determined by
the formula
$R = \frac{V^2}{P}$.
For the 25 W bulb (denoted by subscript 1):
$$
R_1 = \frac{(220)^2}{25}.
$$
For the 100 W bulb (denoted by subscript 2):
$$
R_2 = \frac{(220)^2}{100}.
$$
Step 2: Calculate the Total Resistance in Series
When bulbs (resistors) are connected in series, the total resistance
$R_{\text{total}}$ is the sum of the individual resistances:
$$
R_{\text{total}} = R_1 + R_2.
$$
Step 3: Determine the Current through the Series Connection
Because both bulbs are in series, the same current $I$ flows through each.
The current is given by:
$$
I = \frac{V}{R_{\text{total}}},
$$
where $V = 220\,\text{V}$ is the supply voltage.
Step 4: Calculate the Power Dissipated by Each Bulb
Power dissipated by a resistor $R$ in a series circuit is found using
$P = I^2 R$.
For the 25 W bulb:
$$
P_1 = I^2 \, R_1.
$$
For the 100 W bulb:
$$
P_2 = I^2 \, R_2.
$$
Substituting the expressions for $R_1, R_2,$ and $I$ yields:
$$
P_1 = \left( \frac{220}{R_1 + R_2} \right)^2 \times R_1,
\quad
P_2 = \left( \frac{220}{R_1 + R_2} \right)^2 \times R_2.
$$
Step 5: Numerical Evaluation
Upon carrying out the calculation, we find:
$$
P_1 = 16\,\text{W}, \quad P_2 = 4\,\text{W}.
$$
Hence, the 25 W bulb (with the larger resistance) dissipates 16 W, and the
100 W bulb dissipates 4 W in the series arrangement.
Answer
The correct distribution of power is:
$$
P_1 = 16\,\text{W} \quad \text{and} \quad P_2 = 4\,\text{W}.
$$
