Power Consumption Analysis for Series Connected Bulbs

This problem involves understanding how the power consumed by electric bulbs changes when they are connected in series, especially when their ratings differ. We are given two bulbs with different power ratings but the same voltage rating, connected in series to a supply voltage equal to their rated voltage. We need to find the actual power consumed by the 100 W bulb under these specific conditions.

Key Electrical Concepts for Series Circuits

To solve this, we need to use the relationship between power ($P$), voltage ($V$), and resistance ($R$). The fundamental formula is $P = \frac{V^2}{R}$. From this, we can derive the formula for resistance: $R = \frac{V^2}{P}$. When components are connected in series, the same current flows through each component, and the total resistance is the sum of individual resistances. The power consumed by a component in a series circuit can be calculated using $P = I^2 R$, where $I$ is the current flowing through it.

Step 1: Calculate the Resistance of the 100 W Bulb

The first bulb is rated at $220$ V and $100$ W. We can calculate its resistance ($R_1$) using the formula $R = \frac{V^2}{P}$.

• Rated Voltage ($V_{rated}$) = $220$ V
• Rated Power ($P_{rated1}$) = $100$ W
• Resistance ($R_1$) = $\frac{V_{rated}^2}{P_{rated1}} = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} \Omega = 484 \Omega$

Step 2: Calculate the Resistance of the 50 W Bulb

Similarly, the second bulb is rated at $220$ V and $50$ W. We calculate its resistance ($R_2$).

• Rated Voltage ($V_{rated}$) = $220$ V
• Rated Power ($P_{rated2}$) = $50$ W
• Resistance ($R_2$) = $\frac{V_{rated}^2}{P_{rated2}} = \frac{(220 \text{ V})^2}{50 \text{ W}} = \frac{48400}{50} \Omega = 968 \Omega$

Note that the bulb with lower power rating (50 W) has a higher resistance ($968 \Omega$) compared to the bulb with higher power rating (100 W, $484 \Omega$), when both are designed for the same voltage.

Step 3: Determine the Total Resistance in the Series Circuit

When the two bulbs are connected in series across a $220$ V supply, their resistances add up.

• Total Resistance ($R_{total}$) = $R_1 + R_2$
• $R_{total} = 484 \Omega + 968 \Omega = 1452 \Omega$

Step 4: Calculate the Current Flowing Through the Series Circuit

The current ($I$) flowing through the series circuit can be found using Ohm's Law, $I = \frac{V}{R}$, where $V$ is the supply voltage.

• Supply Voltage ($V_{supply}$) = $220$ V
• Total Resistance ($R_{total}$) = $1452 \Omega$
• Current ($I$) = $\frac{V_{supply}}{R_{total}} = \frac{220 \text{ V}}{1452 \Omega}$

To simplify the fraction for current:

• $I = \frac{220}{1452} \text{ A} = \frac{22 \times 10}{22 \times 66} \text{ A} = \frac{10}{66} \text{ A} = \frac{5}{33} \text{ A}$

So, the current flowing through both bulbs in series is $\frac{5}{33}$ A.

Step 5: Calculate the Power Consumed by the 100 W Bulb

Now we can calculate the actual power consumed by the 100 W rated bulb (which has resistance $R_1 = 484 \Omega$) using the formula $P = I^2 R$.

• Current ($I$) = $\frac{5}{33}$ A
• Resistance of the 100 W bulb ($R_1$) = $484 \Omega$
• Power Consumed ($P_{consumed1}$) = $I^2 R_1 = \left(\frac{5}{33} \text{ A}\right)^2 \times 484 \Omega$
• $P_{consumed1} = \frac{25}{1089} \times 484$ W

To simplify this calculation:

• We know $33^2 = 1089$ and $22^2 = 484$.
• $P_{consumed1} = \frac{25}{33 \times 33} \times (22 \times 22)$ W
• $P_{consumed1} = \frac{25}{(3 \times 11) \times (3 \times 11)} \times (2 \times 11) \times (2 \times 11)$ W
• $P_{consumed1} = \frac{25 \times (2 \times 2) \times (11 \times 11)}{(3 \times 3) \times (11 \times 11)}$ W
• $P_{consumed1} = \frac{25 \times 4 \times 121}{9 \times 121}$ W
• Cancel out the common factor $121$:
• $P_{consumed1} = \frac{25 \times 4}{9}$ W
• $P_{consumed1} = \frac{100}{9}$ W

Conclusion: Power Consumed by the 100 W Bulb

When the $220$ V, $100$ W bulb and the $220$ V, $50$ W bulb are connected in series across a $220$ V supply, the power consumed by the $100$ W bulb is $\frac{100}{9}$ W. This is significantly less than its rated power, which is expected in a series circuit where the current is limited by the total resistance, and the power dissipation is proportional to the resistance ($P = I^2 R$). The bulb with higher resistance (the 50 W rated bulb) will consume more power in this series connection.

Comparing this result with the given options, the calculated power matches the first option.